Question

A voltaic cell consists of an Mn/Mn²⁺ half-cell and a Pb/Pb²⁺ half-cell. Calculate [Pb²⁺] when [Mn²⁺] is 2.7 M and E_cell = 0.35 V.

Answer 1

anode reaction: oxidation takes place

Mn(s) -------------------------> Mn+2 (aq) + 2e-   ,   E⁰_Mn+2/Mn = - 1.18 V

cathode reaction : reduction takes palce

Pb+2(aq) + 2e- -----------------------------> Pb(s) , E⁰_Pb+2/Pb = -0.13V

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net reaction: Mn(s) +Pb+2(aq) -------------------------> Mn+2 (aq) + Pb(s)

E⁰_cell= E⁰_cathode- E⁰_anode

E⁰_cell= E⁰_Pb+2/Pb - E⁰_Mn+2/Mn

          = -0.13 - (-1.18)

          = 1.05 V

nernest equation

Ecell = E⁰_cell -2.303RT/nF* log [Mn+2]/[Pb+2]

Here R= universal gas constant 8.314 J/K mol

T = absolute temperature =25(0C)= 298k

F= faraday = 96500 Coloumb/mol

     n   = no of moles of electrons are transfered =2

2.303RT/F= 0.0591

Ecell = E⁰_cell -(0.0591/n)* log [Mn+2]/[Pb+2]

0.35 = 1.05 - (0.059 x1/2) *log [2.7/Pb+2]

[Pb+2] = 5.64 x 10^-24

[Pb+2] = 5.64 x 10^-24 M