Question 1

Given at 298 K: E^ _ Fe^ 2+ /Fe = X Volt; E^ _ Fe^ 3+ /Fe = Y Volt. The E^ _ Fe^ 3+ /Fe^ 2+ in Volt at 298 K is given by:

Accepted Answer

The half-reactions and their standard Gibbs free energy changes are given by: Fe^ 2+ + 2e^ - Fe G^ _ 1 = -2FX Fe^ 3+ + 3e^ - Fe G^ _ 2 = -3FY The required half-reaction is: Fe^ 3+ + e^ - Fe^ 2+ G^ _ 3 = -1FE^ _ Fe^ 3+ /Fe^ 2+ This reaction is obtained by subtracting the first reaction from the second reaction: G^ _ 3 = G^ _ 2 - G^ _ 1 -FE^ _ Fe^ 3+ /Fe^ 2+ = -3FY - (-2FX) -FE^ _ Fe^ 3+ /Fe^ 2+ = -3FY + 2FX E^ _ Fe^ 3+ /Fe^ 2+ = 3Y - 2X Answer: 3Y-2X

Question 2

For a general redox reaction Anode : Red _1 Ox _1^ n_1^+ + n_1 e^- Cathode : Ox _2 + n_2 e^- Red _2^ n_2^- Which of the following statement is incorrect ?

Accepted Answer

For the given redox reaction, the half-cell reactions are: Anode: Red _1 Ox _1^ n_1^+ + n_1 e^- Cathode: Ox _2 + n_2 e^- Red _2^ n_2^- To obtain the overall balanced chemical equation, we multiply the anode reaction by n_2 and the cathode reaction by n_1 so that the number of electrons lost equals the number of electrons gained: n_2 Red _1 n_2 Ox _1^ n_1^+ + n_1 n_2 e^- n_1 Ox _2 + n_1 n_2 e^- n_1 Red _2^ n_2^- Adding these gives the overall reaction: n_2 Red _1 + n_1 Ox _2 n_2 Ox _1^ n_1^+ + n_

Question 3

Consider the following data. Electrolyte ^ _m (S cm^2 mol^ -1 ) BaCl _2 x_1 H _2 SO _4 x_2 HCl x_3 BaSO _4 is sparingly soluble in water. If the conductivity of the saturated BaSO _4 solution is x S c

Accepted Answer

Using Kohlrausch's law of independent migration of ions, the limiting molar conductivity of BaSO _4 is given by: ^ _m( BaSO _4) = ^ _m( BaCl _2) + ^ _m( H _2 SO _4) - 2 ^ _m( HCl ) Substituting the given values: ^ _m( BaSO _4) = x_1 + x_2 - 2x_3 For a sparingly soluble salt, the solubility S (in mol L^ -1 ) is related to its conductivity (in S cm^ -1 ) and molar conductivity _m (in S cm^2 mol^ -1 ) by the relation: _m = 1000 S Given that = x and _m = ^ _m (which implies complete dissociation, me

Question 4

One half cell in a voltaic cell is constructed by dipping silver rod in AgNO_3 solution of unknown concentration, other half cell is Zn rod dipped in 1 molar solution of ZnSO_4. A voltage of 1.60 V is

Accepted Answer

The cell reaction is Zn(s) + 2Ag^+(aq) Zn^ 2+ (aq) + 2Ag(s) The standard cell potential is given by: E^ _ cell = E^ _ cathode - E^ _ anode = 0.80 - (-0.76) = 1.56 V Using the Nernst equation: E_ cell = E^ _ cell - 0.059 n [Zn^ 2+ ] [Ag^+]^2 Substituting the given values (n = 2, [Zn^ 2+ ] = 1 M , E_ cell = 1.60 V ): 1.60 = 1.56 - 0.059 2 1 x^2 0.04 = - 0.059 2 (-2 x) 0.04 = 0.059 x x = 0.04 0.059 = 4 5.9 Answer: 4 5.9

Question 5

At 298 K , the molar conductivity of x\% (w/w) MX solution (aqueous) is 123.5 S cm ^2 mol ^ -1 . The conductance of same solution is 1.9 10^ -3 S . The value of x is _______ 10^ -2 . (Given : cell con

Accepted Answer

The conductivity (specific conductance), , is given by the product of conductance and cell constant: = G G^* = 1.9 10^ -3 S 1.3 cm ^ -1 = 2.47 10^ -3 S cm ^ -1 The molar conductivity, _m, is related to the conductivity and molarity C by the formula: _m = 1000 C Substituting the given values: 123.5 = 2.47 10^ -3 1000 C 123.5 = 2.47 C C = 2.47 123.5 = 0.02 mol L ^ -1 Molarity C can also be expressed in terms of mass percentage x (w/w), density d, and molar mass M: C = x d 10 M Substituting the kno

Question 6

An electrochemical cell, consist of the following two redox couples, M^ x+ (aq)/M(s) [E_ red ^ =+0.15 V ] and Fe^ 3+ (aq)/Fe(s) [E_ red ^ =-0.036 V ]. The cell EMF (E_ cell ) is recorded to be 0.2057

Accepted Answer

Given standard reduction potentials: E^ _ M^ x+ /M = +0.15 V E^ _ Fe^ 3+ /Fe = -0.036 V Since E^ _ M^ x+ /M > E^ _ Fe^ 3+ /Fe , reduction occurs at the M^ x+ /M electrode (cathode) and oxidation occurs at the Fe^ 3+ /Fe electrode (anode). Cathode reaction: (M^ x+ + x e^ - M) 3 Anode reaction: (Fe Fe^ 3+ + 3 e^ - ) x Overall balanced cell reaction: 3 M^ x+ + x Fe 3 M + x Fe^ 3+ The number of electrons transferred in the balanced reaction is n = 3x. Standard cell potential E^ _ cell is: E^ _ cell

Question 7

Consider the following two half-cell reactions along with the standard reduction potential given: CO_2 + 6H^+ + 6e^- CH_3 OH + H_2 O E°_ red = 0.02 V 1 2 O_2 + 2H^+ + 2e^- H_2 O E°_ red = 1.23 V A fue

Accepted Answer

The half-cell reactions for the fuel cell are: Anode (Oxidation): CH_3OH + H_2O CO_2 + 6H^+ + 6e^- E^ _ ox = -0.02 V Cathode (Reduction): 3 2 O_2 + 6H^+ + 6e^- 3H_2O E^ _ red = 1.23 V The overall cell reaction is: CH_3OH + 3 2 O_2 CO_2 + 2H_2O The standard cell potential is: E^ _ cell = E^ _ cathode - E^ _ anode = 1.23 - 0.02 = 1.21 V The number of electrons transferred per mole of CH_3OH is n = 6. The maximum electrical work that can be obtained from the cell is given by the decrease in Gibbs f

Question 8

An electrochemical cell is constructed using half cells in the direction of spontaneous change Fe(OH)_2(s) + 2e^- Fe(s) + 2OH^-(aq) E^ = -0.88 V and AgBr(s) + e^- Ag(s) + Br^-(aq) E^ = +0.07 V Which o

Accepted Answer

For a spontaneous reaction, the cell potential E^ _ cell must be positive. The half-cell with the higher standard reduction potential acts as the cathode (reduction), and the one with the lower standard reduction potential acts as the anode (oxidation). Given standard reduction potentials: E^ _ AgBr/Ag = +0.07 V E^ _ Fe(OH)_2/Fe = -0.88 V Since +0.07 V > -0.88 V, AgBr undergoes reduction at the cathode and Fe undergoes oxidation at the anode. Cathode reaction: 2AgBr(s) + 2e^ - 2Ag(s) + 2Br^ - (a

Question 9

For strong electrolyte _ m increases slowly with dilution and can be represented by the equation _ m = _ m ^ - Ac ^ 1 / 2 Molar conductivity values of the solutions of strong electrolyte AB at 18^ C a

Accepted Answer

The equation _m = _m^0 - Ac^ 1/2 is linear in c^ 1/2 . Taking two data points: (c=0.04, c =0.2, _m=96.1) and (c=0.25, c =0.5, _m=94.9). Slope = 94.9 - 96.1 0.5 - 0.2 = -1.2 0.3 = -4.0 Since slope = -A, the constant A = 4 S cm² mol⁻¹/(mol/L)^(1/2)

Question 10

A volume of x\, mL of 5\, M \, NaHCO_3 solution was mixed with 10\, mL of 2\, M \, H_2CO_3 solution to make an electrolytic buffer. If the same buffer was used in the following electrochemical cell to

Accepted Answer

We have considered E^ _ [Sn(OH)_6]^ 2- /HSnO_2^- = -0.9 V Pt | HSnO_2^-(aq), [Sn(OH)_6]^ 2- (aq), OH^-(aq) | Bi_2O_3(s) | Bi(s) | 0.5M 0.05M E^ _ cell = +0.9 - 0.44 = 0.46 V Oxidation Half : HSnO_2^- + H_2O + 3OH^- [Sn(OH)_6]^ 2- + 2e^- Reduction Half : Bi_2O_3 + 3H_2O + 6e^- 2Bi + 6OH^- --------------------------------------- 3HSnO_2^-(aq) + Bi_2O_3(s) + 6H_2O + 3OH^-(aq) 3[Sn(OH)_6]^ 2- (aq) + 2Bi(s) E_ cell = E^ _ cell - 0.059 6 (0.5)^3 (0.05)^3 [OH^-]^3 0.2353 = 0.46 - 0.059 6 3 [ 10 [OH^-]

Question 11

Consider the following redox reaction taking place in acidic medium BH _ 4 ^ - (a q)+ ClO _ 3 ^ - (a q) H _ 2 BO _ 3 ^ - (a q)+ Cl ^ - (a q) If the Nernst equation for the above balanced reaction is E

Accepted Answer

3BH_4^- + 4ClO_3^- 4Cl^- + 3H_2BO_3^- + 3H_2O n-factor = 8 moles = 3 n = 3 8 = 24

Question 12

Molar conductivity of a weak acid HQ of concentration 0.18 M was found to be 1 / 30 of the molar conductivity of another weak acid HZ with concentration of 0.02 M. If ^ Q ^ - happened to be equal with

Accepted Answer

For weak acids with ^ Q^- = ^ Z^-, the molar conductivity is _m = ^ . From _m^Q = 1 30 _m^Z, we get _Q = _Z 30 . Using K_a = ^2 C for weak acids: K_a(HQ) = _Q^2 0.18 = _Z^2 0.18 900 and K_a(HZ) = _Z^2 0.02. Therefore K_a(HZ) K_a(HQ) = 0.02 900 0.18 = 100. Thus pK_a(HQ) - pK_a(HZ) = (100) = 2.

Question 13

Electricity is passed through an acidic solution of Cu ^ 2+ till all the Cu ^ 2+ was exhausted, leading to the deposition of 300 mg of Cu metal. However, a current of 600 mA was continued to pass thro

Accepted Answer

Eq of Cu = Eq of O_2 300 10^ -3 2 63.54 = n_ O_2 4 2.36 10^ -3 = n_ O_2 When current is further passed n_ O_2 4 = 600 28 60 96500 1000 n_ O_2 = 2.611 10^ -3 Total O_2 released = [10^ -3 (2.36 + 2.611)] 22400 ml = 111.35 ml

Question 14

Consider the above electrochemical cell where a metal electrode (M) is undergoing redox reaction by forming M ^ + ( M M ^ + + e ^ - ). The cation M ^ + is present in two different concentrations c_ 1

Accepted Answer

The given setup is a concentration cell with identical electrodes (M) and different electrolyte concentrations (c_1 and c_2). For a concentration cell, the standard cell potential E^0_ cell = 0. The cell reactions are: Anode: M(s) M^+(c_ anode ) + e^- Cathode: M^+(c_ cathode ) + e^- M(s) Net reaction: M^+(c_ cathode ) M^+(c_ anode ) Using the Nernst equation at 298 K : E_ cell = E^0_ cell - 0.0591 n c_ anode c_ cathode E_ cell = 0 - 0.0591 c_ anode c_ cathode For a positive cell potential (E_ ce

Question 15

In the given electrochemical cell, Ag ( s )| AgCl ( s )| FeCl _ 2 ( aq ), FeCl _ 3 ( aq ) Pt ( s ) at 298 K, the cell potential ( E _ cell ) will increase when : A. Concentration of Fe ^ 2+ is increas

Accepted Answer

The cell Ag(s)|AgCl(s)|FeCl₂(aq), FeCl₃(aq)|Pt(s) has the reaction: Ag(s) + Cl⁻ + Fe³⁺ → AgCl(s) + Fe²⁺ Using Nernst equation: E_ cell = E°_ cell - 0.059 n Q where Q = [Fe^ 2+ ] [Fe^ 3+ ][Cl^-] For E_ cell to increase, Q must decrease. Analyzing each condition: A. Increase [Fe²⁺] increases Q, decreases E_ cell - false B. Decrease [Fe³⁺] increases Q (smaller denominator), decreases E_ cell - false C. Decrease [Fe²⁺] decreases Q, increases E_ cell - true D. Increase [Fe³⁺] decreases Q, increases E

Question 16

Consider the following reduction processes : Al ^ 3+ +3 e ^ - Al ( s ), E ^ 0 =-1.66 ~V Fe ^ 3+ + e ^ - Fe ^ 2+ , E ^ 0 =+0.77 ~V Co ^ 3+ + e ^ - Co ^ 2+ , E ^ 0 =+1.81 ~V Cr ^ 3+ +3 e ^ - Cr ( s ), E

Accepted Answer

Reducing agent strength is determined by oxidation potential (opposite sign of reduction potential). Stronger reducing agents have more negative oxidation potentials (easier to oxidize). Oxidation potentials: Al(s) → Al³⁺ has E° = +1.66 V (strongest reducing agent); Cr(s) → Cr³⁺ has E° = +0.74 V; Fe²⁺ → Fe³⁺ has E° = -0.77 V; Co²⁺ → Co³⁺ has E° = -1.81 V (weakest reducing agent). Order of decreasing reducing agent tendency: Al > Cr > Fe²⁺ > Co²⁺.

Question 17

Consider the following electrochemical cell : Pt | O _ 2 ( ~g )(1 bar ) | HCl ( aq ) \| M ^ 2+ ( aq , 1.0 M ) M ( ~s ) The pH above which, oxygen gas would start to evolve at anode is \_\_\_\_ (neares

Accepted Answer

In this electrochemical cell, at the anode oxygen evolution occurs from water oxidation. Using Nernst equation for the O₂/H₂O couple: E_ anode = E°_ O_2/H_2O - 0.059 4 1 [H^+]^4 = 1.23 - 0.059 pH At the cathode: E_ cathode = 0.994 V (standard potential for M²⁺/M reduction) For oxygen to start evolving, the cell potential becomes zero: E_ cell = E_ cathode - E_ anode = 0 0.994 = 1.23 - 0.059 pH 0.059 pH = 0.236 pH = 4.0

Question 18

Consider the following electrochemical cell at 298 K Pt | HSnO _ 2 ^ - ( aq ) | Sn ( OH )_ 6 ^ 2- ( aq ) | OH ^ - ( aq ) | Bi _ 2 O _ 3 ( ~s ) Bi ( s ). If the reaction quotient at a given time is 10^

Accepted Answer

The electrochemical cell has cathode: Bi_2O_3 + 3H_2O + 6e^- 2Bi + 6OH^- with E° = -0.44 V and anode: Sn(OH)_6^ 2- - 2e^- HSnO_2^- + H_2O + OH^- with E° = -0.90 V. The standard cell EMF is E°_ cell = -0.44 - (-0.90) = 0.46 V. Using the Nernst equation with Q = 10^6 and n = 6 electrons transferred: E_ cell = 0.46 - 0.059 6 (10^6) = 0.46 - 0.059 6 6 = 0.46 - 0.059 = 0.401 V. Expressing as x 10^ -1 V: 0.401 = 4.01 10^ -1 V, so x = 4 (nearest integer).

Question 19

For a closed circuit Daniell cell, which of the following plots is the accurate one at a given temperature?

Accepted Answer

The standard cell potential E^ _ cell is a thermodynamic property defined for a cell under standard conditions (concentrations of all species at 1 M , pressure at 1 bar , and a specific temperature). For a Daniell cell, the reaction is Zn(s) + Cu^ 2+ (aq) Zn^ 2+ (aq) + Cu(s). The standard cell potential is given by E^ _ cell = E^ _ Cu^ 2+ /Cu - E^ _ Zn^ 2+ /Zn . Since E^ _ cell depends only on the nature of the electrodes, the nature of the electrolytes, and the temperature, it remains constant

Question 20

MX is a sparingly soluble salt that follows the given solubility equilibrium at 298 K. MX ( s ) M ^ + ( aq )+ X ^ - ( aq ) ; K _ sp =10^ -10 If the standard reduction potential for M ^ + ( aq ) + e ^

Accepted Answer

The standard reduction potential for a metal/metal insoluble salt electrode is related to the standard reduction potential of the metal ion and the solubility product K_ sp of the salt. The half-cell reaction for the X^-/MX(s)/M electrode is: MX(s) + e^- M(s) + X^-(aq). This can be viewed as a combination of two processes: 1. MX(s) M^+(aq) + X^-(aq) with equilibrium constant K_ sp = 10^ -10 2. M^+(aq) + e^- M(s) with E^ _ M^+/M = 0.79 V The relation between the standard potentials is given by: E

Electrochemistry — Previous Year Questions & Solutions