Give the Nernst equation for the electrode potential $E_{(M^{n+} \mid M)}$.
The reduction half-reaction for a metal electrode is given by: $M^{n+}(aq) + ne^- \rightarrow M(s)$.
According to the Nernst equation,the electrode potential at any concentration is given by:
$E_{(M^{n+} \mid M)} = E^{\circ}_{(M^{n+} \mid M)} - \frac{RT}{nF} \ln \frac{[M(s)]}{[M^{n+}(aq)]}$.
Since the concentration of a pure solid is taken as unity $([M(s)] = 1)$,the equation simplifies to:
$E_{(M^{n+} \mid M)} = E^{\circ}_{(M^{n+} \mid M)} - \frac{RT}{nF} \ln \frac{1}{[M^{n+}(aq)]}$.
Alternatively,using base $10$ logarithm at $298 \ K$:
$E_{(M^{n+} \mid M)} = E^{\circ}_{(M^{n+} \mid M)} - \frac{0.0591}{n} \log \frac{1}{[M^{n+}(aq)]}$.
According to the Nernst equation,the electrode potential at any concentration is given by:
$E_{(M^{n+} \mid M)} = E^{\circ}_{(M^{n+} \mid M)} - \frac{RT}{nF} \ln \frac{[M(s)]}{[M^{n+}(aq)]}$.
Since the concentration of a pure solid is taken as unity $([M(s)] = 1)$,the equation simplifies to:
$E_{(M^{n+} \mid M)} = E^{\circ}_{(M^{n+} \mid M)} - \frac{RT}{nF} \ln \frac{1}{[M^{n+}(aq)]}$.
Alternatively,using base $10$ logarithm at $298 \ K$:
$E_{(M^{n+} \mid M)} = E^{\circ}_{(M^{n+} \mid M)} - \frac{0.0591}{n} \log \frac{1}{[M^{n+}(aq)]}$.
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Similar Questions
Calculate the cell potential at $298 \ K$ for the following cells:
$(a)$ $Cd \mid Cd^{2+}(0.02 \ M) \parallel H^{+}(1 \ M) \mid H_{2(g)}(1 \ bar) \mid Pt$ $\left[ E_{Cd^{2+} \mid Cd}^0 = -0.40 \ V \right]$
$(b)$ $Al \mid Al^{3+}(0.25 \ M) \parallel Zn^{2+}(0.15 \ M) \mid Zn_{(s)}$ $\left[ E_{Al^{3+} \mid Al}^0 = -1.66 \ V, E_{Zn^{2+} \mid Zn}^0 = -0.76 \ V \right]$
$(a)$ $Cd \mid Cd^{2+}(0.02 \ M) \parallel H^{+}(1 \ M) \mid H_{2(g)}(1 \ bar) \mid Pt$ $\left[ E_{Cd^{2+} \mid Cd}^0 = -0.40 \ V \right]$
$(b)$ $Al \mid Al^{3+}(0.25 \ M) \parallel Zn^{2+}(0.15 \ M) \mid Zn_{(s)}$ $\left[ E_{Al^{3+} \mid Al}^0 = -1.66 \ V, E_{Zn^{2+} \mid Zn}^0 = -0.76 \ V \right]$
Medium
View SolutionAt $298 \ K$,the following reaction takes place for a hydrogen electrode:
$H^{+}_{(aq)} + e^{-} \longrightarrow \frac{1}{2} H_2(1 \ bar)$
The solution $pH$ is $10.0$. What is the hydrogen electrode potential in volts?
$\left(\frac{2.303 RT}{F} = 0.06 \ V\right)$
$H^{+}_{(aq)} + e^{-} \longrightarrow \frac{1}{2} H_2(1 \ bar)$
The solution $pH$ is $10.0$. What is the hydrogen electrode potential in volts?
$\left(\frac{2.303 RT}{F} = 0.06 \ V\right)$
For the reduction of silver ions by copper metal,the standard cell potential at $25\,^{\circ}C$ is $0.46\,V$. What is the value of the standard Gibbs energy $(\Delta G^{\circ})$? $(F = 96500\,C\,mol^{-1})$
Medium
View SolutionWhat happens to the $emf$ of the cell $Zn_{(s)} | Zn^{+2} (1 \ M) || Ag^{+1} (1 \ M) | Ag_{(s)}$ if the concentration of $Ag^{+1}$ decreases to $0.1 \ M$?
$1 \ F$ electricity was passed through $Cu^{2+} (1.5 \ M, 1 \ L) / Cu$ and $0.1 \ F$ was passed through $Ag^{+} (0.2 \ M, 1 \ L) / Ag$ electrolytic cells. After this,the two cells were connected to make an electrochemical cell. The $emf$ of the cell thus formed at $298 \ K$ is:
Given: $E^0_{Cu^{2+} / Cu} = 0.34 \ V$,$E^0_{Ag^{+} / Ag} = 0.8 \ V$,$\frac{2.303 \ RT}{F} = 0.06 \ V$ (in $V$)
Given: $E^0_{Cu^{2+} / Cu} = 0.34 \ V$,$E^0_{Ag^{+} / Ag} = 0.8 \ V$,$\frac{2.303 \ RT}{F} = 0.06 \ V$ (in $V$)
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