A voltaic cell consists of an Al/Al3+ half-cell and a Cd/Cd2+ half-cell. Calculate {Cd2+} when {Al3+}...

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A voltaic cell consists of an Al/Al3+ half-cell and a Cd/Cd2+ half-cell. Calculate {Cd2+} when {Al3+}...

A voltaic cell consists of an Al/Al3+ half-cell and a Cd/Cd2+ half-cell. Calculate {Cd2+} when {Al3+} = 0.306 M and Ecell = 1.27 V. Use reduction potential values of Al3+ = -1.66 V and for Cd2+ = -0.40 V.

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Answer 1

Answer #1

The reactions are Al+3--------> Al++3e- Eo= 1.66V (1) and Cd+2 +2e- - -----> Cd E0=-0.40V (2)

Multiply Eq.1 with 2 and Eq.2 with 3 gives

2Al+3 + 3Cd ----------2Al + 3Cd+2 Eo= 1.66-0.4= 1.26V

E = EO- (0.0592/n)* loqQ, n = number of eletrons and Q= [Cd+2]³/ [Al+3^]2

1.27= 1.26 -0.0592/6* logQ

Q= 0.097=[ Cd+2]3/ (0.306⁾², [Cd+2]³= 0.097*0.306*0.306

[Cd+2]= 0.2086

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Answer 2

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