Which of the following is the correct expression for the electrode potential of a cell?
- A$E = E^o - \frac{RT}{nF} \ln \frac{[\text{product}]}{[\text{reactant}]}$
- B$E = E^o + \frac{RT}{F} \ln \frac{[\text{product}]}{[\text{reactant}]}$
- C$E = E^o - \frac{RT}{nF} \ln \frac{[\text{reactant}]}{[\text{product}]}$
- D$E = - \frac{RT}{F} \ln \frac{[\text{product}]}{[\text{reactant}]}$
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Similar Questions
For the following electrochemical cell at $298 \ K$,
$Pt_{(s)} \mid H_2(g, 1 \ bar) \mid H^{+}(aq, 1 \ M) \parallel M^{4+}_{(aq)}, M^{2+}_{(aq)} \mid Pt_{(s)}$
$E_{\text{cell}} = 0.092 \ V$ when $\frac{[M^{2+}_{(aq)}]}{[M^{4+}_{(aq)}]} = 10^x$
Given : $E^0_{M^{4+}/M^{2+}} = 0.151 \ V$; $2.303 \frac{RT}{F} = 0.059 \ V$
The value of $x$ is
$Pt_{(s)} \mid H_2(g, 1 \ bar) \mid H^{+}(aq, 1 \ M) \parallel M^{4+}_{(aq)}, M^{2+}_{(aq)} \mid Pt_{(s)}$
$E_{\text{cell}} = 0.092 \ V$ when $\frac{[M^{2+}_{(aq)}]}{[M^{4+}_{(aq)}]} = 10^x$
Given : $E^0_{M^{4+}/M^{2+}} = 0.151 \ V$; $2.303 \frac{RT}{F} = 0.059 \ V$
The value of $x$ is
Which one of the following has a potential more than zero?
What is the reduction electrode potential $E$ of a $0.1 \, M$ solution of $M^{+}$ ions,given that the standard reduction potential $E^o_{RP} = -2.36 \, V$?
Medium
View SolutionCalculate the $pH$ of an $HCl$ solution at $298 \ K$ for the following cell:
$Pt_{(s)} \mid H_2 \ (1 \ bar) \mid HCl \ (xM) \parallel Ag^+ \ (0.01 \ M) \mid Ag_{(s)}$
Given that the standard cell potential $E^\circ_{cell} = 1.05 \ V$. (in $.73$)
$Pt_{(s)} \mid H_2 \ (1 \ bar) \mid HCl \ (xM) \parallel Ag^+ \ (0.01 \ M) \mid Ag_{(s)}$
Given that the standard cell potential $E^\circ_{cell} = 1.05 \ V$. (in $.73$)
Difficult
View SolutionConsider the cell
$Pt_{(s)} \mid H_2(g, 1\,atm) \mid H^{+}(aq, 1\,M) \parallel Fe^{3+}_{(aq)}, Fe^{2+}_{(aq)} \mid Pt_{(s)}$
When the potential of the cell is $0.712\,V$ at $298\,K$,the ratio $[Fe^{2+}] / [Fe^{3+}]$ is $.......$ (Nearest integer).
Given: $Fe^{3+} + e^- \longrightarrow Fe^{2+}$,$E^{\circ}_{Fe^{3+}/Fe^{2+}} = 0.771\,V$
$\frac{2.303 RT}{F} = 0.06\,V$
$Pt_{(s)} \mid H_2(g, 1\,atm) \mid H^{+}(aq, 1\,M) \parallel Fe^{3+}_{(aq)}, Fe^{2+}_{(aq)} \mid Pt_{(s)}$
When the potential of the cell is $0.712\,V$ at $298\,K$,the ratio $[Fe^{2+}] / [Fe^{3+}]$ is $.......$ (Nearest integer).
Given: $Fe^{3+} + e^- \longrightarrow Fe^{2+}$,$E^{\circ}_{Fe^{3+}/Fe^{2+}} = 0.771\,V$
$\frac{2.303 RT}{F} = 0.06\,V$
DifficultJEE MAIN 2023
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