To find the standard potential of M ^ 3 + / M electrode, the following cell is constituted: Pt / M / M ^ 3 + (0 .001 mol L ^ - 1 ) / Ag ^ + (0 .01 mol L ^ - 1 ) / Ag The emf of the cell is found to be 0 . 421volt at

Correct answer: C. 0 . 32volt E _ cell = E ^ o cell - ((0 . 0591)/(n))logQ ∴ Q = ( M ^ 3 + ) ( Ag ^ + ) ^ 3 = 0 .001 ( 0 .01 ) ^ 3 = 10 ^ 3 E _ cell = E ^ o cell - ((0 . 0591)/(3))log 10 ^ 3 E _ cell = E ^ o cell - ((0 . 0591)/(3)) × 3 = 0 .421 E ^ o cell - 0 .0591 = 0 . 421 E ^ o cell = 0 . 48V = E ^ o Ag ^ + / Ag - E ^ o M ^ 3 + / M E ^ o M ^ 3 + / M = 0 . 80 - 0 . 48 = 0 . 32V

JEE Main 20179 AprilOnline

Exam: JEE Main
Year: 2017
Date: 9 April
Mode: Online
Subject: Chemistry
Chapter: Electrochemistry

JEE Main 2017 · 9 April · Online · Chemistry · Electrochemistry

To find the standard potential of M ^ 3 + / M electrode, the...

To find the standard potential of

{M}^{3 +} / M

electrode, the following cell is constituted:

Pt / M / {M}^{3 +}\left(0 .001 mol{L}^{- 1}\right) / {Ag}^{+}\left(0 .01 mol{L}^{- 1}\right) / Ag

The emf of the cell is found to be

0 . 421volt

298K

. The standard potential of half-reaction

{M}^{3 +} + 3{e}^{-} \to M

298K

will be:

(Given:

{E}_{\dfrac{{Ag}^{+}}{Ag}}^{⊝}

298K = 0 . 80volt

)

Options

A $0 . 38volt$
B $1 . 28volt$
C $0 . 32volt$
D $0 . 66volt$

Correct answer

C. $0 . 32volt$

Step-by-step solution

{E}_{cell} = {E}^{o}cell - \left(\dfrac{0 . 0591}{n}\right)logQ

∴ Q = \dfrac{\left({M}^{3 +}\right)}{{\left({Ag}^{+}\right)}^{3}} = \dfrac{0 .001}{{( 0 .01 )}^{3}} = {10}^{3}

{E}_{cell} = {E}^{o}cell - \left(\dfrac{0 . 0591}{3}\right)log{10}^{3}

{E}_{cell} = {E}^{o}cell - \left(\dfrac{0 . 0591}{3}\right) \times 3 = 0 .421 {E}^{o}cell - 0 .0591 = 0 . 421{E}^{o}cell = 0 . 48V = {E}^{o}{Ag}^{+} / Ag - {E}^{o}{M}^{3 +} / M{E}^{o}{M}^{3 +} / M = 0 . 80 - 0 . 48 = 0 . 32V

Practice Electrochemistry on Quantrex Academy →

More from Electrochemistry

At 300 K, the molar conductivities of the aqueous solutions of three salts at two… JEE Advanced 2026 Given at 298 K: E^ _ Fe^ 2+ /Fe = X Volt; E^ _ Fe^ 3+ /Fe = Y Volt. The E^ _ Fe^ 3+ /Fe^… JEE Main 2026 For a general redox reaction Anode : Red_1 arrow Ox_1^ n_1^+ + n_1 e^- Cathode : Ox_2 +… JEE Main 2026 Consider the following data. Electrolyte Λ^ ∘ _m (S cm ^2 mol ^ -1 ) BaCl_2 x_1 H_2SO_4… JEE Main 2026 One half cell in a voltaic cell is constructed by dipping silver rod in AgNO_3 solution… JEE Main 2026 At 298 K , the molar conductivity of x\% (w/w) MX solution (aqueous) is 123.5 S cm^2 mol^… JEE Main 2026 An electrochemical cell, consist of the following two redox couples, M^ x+ (aq)/M(s) [E_… JEE Main 2026 Consider the following two half-cell reactions along with the standard reduction… JEE Main 2026 Full Electrochemistry list All JEE Main PYQs