Calculate {Cd2+} when {Al3+} = 0.57 M and Ecell = 1.22 V.

Question

A voltaic cell consists of an Al/Al³⁺ half-cell and a Cd/Cd²⁺ half-cell. Calculate {Cd²⁺} when {Al³⁺} = 0.57 M and E_cell = 1.22 V. Use reduction potential values of Al³⁺ = -1.66 V and for Cd²⁺ = -0.40 V.

The correct answer is 0.031 

Answer 1

Cd²⁺ + 2e- ==>Cd(s) reduction half reaction x3

Al(s) ===> Al³⁺ + 3e- oxidation half reaction x2

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3Cd²⁺ ===> 3Cd(s)

2Al(s) ===> 3Al³⁺

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3Cd²⁺ + 2Al(s) ==> 3Cd(s) + 2Al³⁺ overall reaction

Eºcell = -0.40 - (-1.66) = 1.26 V

Ecell = Eº - RT/nF ln Q and Q = [Al³⁺]² / [Cd²⁺]³ = 0.57²/x³

1.22 = 1.26 -(0.0592/6) log Q

-0.04 = -0.0099 log Q

log Q = 4.04

4.04 = log 0.57² - log x³

4.04 = -0.488 - log x³

log x³ = -4.53

x³ = 2.95x10^-5

x = 0.031 M