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Question: For a cell reaction: \(M_{(aq)}^{n +} + ne^{-} \rightarrow M_{(s)}\) the Nernst equation for electro...

For a cell reaction: M(aq)n++ne−→M(s)M_{(aq)}^{n +} + ne^{-} \rightarrow M_{(s)} the Nernst equation for electrode potential at any concentration measured with respect to standard hydrogen electrode is represented as

A

E(Mn+/M)=Eº(Mn+/M)−RTnF1n1[Mn+]E_{(M^{n +}/M)} = Eº_{(M^{n +}/M)} - \frac{RT}{nF}1n\frac{1}{\lbrack M^{n +}\rbrack}

B

E(M/Mn+)=Eº(M/Mn+)−RTnF1n[Mn+][M]E_{(M/M^{n +})} = Eº_{(M/M^{n +})} - \frac{RT}{nF}1n\frac{\lbrack M^{n +}\rbrack}{\lbrack M\rbrack}

C

E(Mn+/M)=Eº(Mn+/M)−RTnF1n1[M]E_{(M^{n +}/M)} = Eº_{(M^{n +}/M)} - \frac{RT}{nF}1n\frac{1}{\lbrack M\rbrack}

D

E(Mn+/M)=Eº(Mn+/M)−RTnF1n[Mn+]E_{(M^{n +}/M)} = Eº_{(M^{n +}/M)} - \frac{RT}{nF}1n\lbrack M^{n +}\rbrack

Answer

E(Mn+/M)=Eº(Mn+/M)−RTnF1n1[Mn+]E_{(M^{n +}/M)} = Eº_{(M^{n +}/M)} - \frac{RT}{nF}1n\frac{1}{\lbrack M^{n +}\rbrack}

Explanation

Solution

E(Mn+/M)=Eº(Mn+/M)−RTnFIn[M][Mn+]E_{(M^{n +}/M)} = Eº_{({M^{n}}^{+}/M)} - \frac{RT}{nF}In\frac{\lbrack M\rbrack}{\lbrack M^{n +}\rbrack}

Since concentration of solid is taken as unity,

E(Mn+/M)=Eº(Mn+/M)−−RTnfln⁡1[Mn+1]E_{(M^{n +}/M)} = Eº_{(M^{n +}/M) -} - \frac{RT}{nf}\ln\frac{1}{\lbrack M^{n + 1}\rbrack}