JEE MainChemistryElectrochemistry
A cell is constructed at 298 K using a standard hydrogen electrode immersed in a 0.1 M solution of a weak acid HA , connected to a silver electrode in a 0.01 M AgNO ₃ solution. The cell representation is: Pt(s) H ₂ (g, 1 bar) HA(aq, 0.1 M) Ag ⁺ (aq, 0.01 M) Ag(s) Given: Acid dissociation constant ( K_ a ) of HA = 10⁻⁵ E^ _ Ag ⁺/ Ag = 0.80 V 2.303RT F = 0.06 V The cell potential E_ cell is _____ 10⁻² V . (Nearest inte
Correct answer
86
Step-by-step solution
First, calculate the H ⁺ ion concentration from the dissociation of the weak acid HA using Ostwald's dilution law: [ H ⁺] = K_ a C [ H ⁺] = 10⁻⁵ 0.1 = 10⁻⁶ = 10⁻³ M The cell reactions are: Anode: 1 2 H ₂ H ⁺ + e⁻ Cathode: Ag ⁺ + e⁻ Ag Overall reaction: 1 2 H ₂ + Ag ⁺ H ⁺ + Ag Here, n = 1 . The standard cell potential is: E^ _ cell = E^ _ cathode - E^ _ anode = 0.80 - 0.00 = 0.80 V The reaction quotient Q is: Q = [ H ⁺] [ Ag ⁺] = 10⁻³ 0.01 = 10⁻¹ Applying the Nernst equation: E_ cell = E^ _ cell - 0.06 n Q E_ cell =