(B) The reduction half-reaction is: $Fe^{3+} + e^{-} \rightarrow Fe^{2+}$Using the Nernst equation:$E_{Fe^{3+} \mid Fe^{2+}} = E^0_{Fe^{3+} \mid Fe^{2

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(B) The reduction half-reaction is: $Fe^{3+} + e^{-} \rightarrow Fe^{2+}$Using the Nernst equation:$E_{Fe^{3+} \mid Fe^{2+}} = E^0_{Fe^{3+} \mid Fe^{2+}} - \frac{2.303 \ RT}{nF} \log \frac{[Fe^{2+}]}{[Fe^{3+}]}$Here,$n = 1$,$[Fe^{2+}] = 2.0 \ M$,$[Fe^{3+}] = 0.02 \ M$,$\frac{2.303 \ RT}{F} = 0.059 \ V$,and $E^0 = 0.771 \ V$.Substituting the values:$E = 0.771 - 0.059 \log \left(\frac{2.0}{0.02}\right)$$E = 0.771 - 0.059 \log(100)$$E = 0.771 - 0.059 \times 2$$E = 0.771 - 0.118 = 0.653 \ V$

Class 12 Chemistry Electrochemistry Nernst equation and ECS

Medium

The reduction potential of a half cell consisting of a $Pt$ electrode immersed in $2.0 \ M \ Fe^{2+}$ and $0.02 \ M \ Fe^{3+}$ solution (in $V$) is. Given: $\left(\frac{2.303 \ RT}{F} = 0.059, E^0_{Fe^{3+} \mid Fe^{2+}} = 0.771 \ V\right)$

TS EAMCET 2023 All TS EAMCET

A
$0.543$
B
$0.653$
C
$0.733$
D
$0.822$

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Nernst equation and ECS

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Explain the equilibrium state in a Daniell cell and derive its equilibrium constant. Given: $E^o_{cell} = 1.1 \ V$.

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The standard electrode potential for $Cu^{+2}/Cu$ is $0.34 \ V$. Calculate the reduction potential at $pH = 14$ for the above couple $V$ $[K_{sp}[Cu(OH)_2] = 1 \times 10^{-19}]$

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The standard emf for the cell $Cd_{(s)}|Cd^{2+}_{(aq)}(1 \ M)||Cu^{2+}_{(aq)}(1 \ M)|Cu_{(s)}$ is $0.74 \ V$. If the concentration of $Cd^{2+}_{(aq)}$ and $Cu^{2+}_{(aq)}$ both decrease by $10$ times at $298 \ K$,calculate the emf of the cell.

MediumMHT CET 2025

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Consider the following cell reaction:
$2Fe_{(s)} + O_{2(g)} + 4H^{+}_{(aq)} \to 2Fe^{2+}_{(aq)} + 2H_2O_{(l)}$; $E^o = 1.67 \ V$
At $[Fe^{2+}] = 10^{-3} \ M$,$p(O_2) = 0.1 \ atm$ and $pH = 3$,the cell potential at $25 \ ^oC$ is .............. $V$.

DifficultAIIMS 2017

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At $298 \ K$,find out the $emf$ for the cell:
$Al_{(s)} | Al^{+3} (0.1 \ M) || Fe^{+2} (0.001 \ M) | Fe_{(s)}$
Given: $E^o_{Al^{+3}/Al} = -1.66 \ V$ and $E^o_{Fe^{+2}/Fe} = -0.44 \ V$.

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The reduction potential of a half cell consisting of a $Pt$ electrode immersed in $2.0 \ M \ Fe^{2+}$ and $0.02 \ M \ Fe^{3+}$ solution (in $V$) is. Given: $\left(\frac{2.303 \ RT}{F} = 0.059, E^0_{Fe^{3+} \mid Fe^{2+}} = 0.771 \ V\right)$

Similar Questions

Explain the equilibrium state in a Daniell cell and derive its equilibrium constant. Given: $E^o_{cell} = 1.1 \ V$.

The standard electrode potential for $Cu^{+2}/Cu$ is $0.34 \ V$. Calculate the reduction potential at $pH = 14$ for the above couple $V$ $[K_{sp}[Cu(OH)_2] = 1 \times 10^{-19}]$

The standard emf for the cell $Cd_{(s)}|Cd^{2+}_{(aq)}(1 \ M)||Cu^{2+}_{(aq)}(1 \ M)|Cu_{(s)}$ is $0.74 \ V$. If the concentration of $Cd^{2+}_{(aq)}$ and $Cu^{2+}_{(aq)}$ both decrease by $10$ times at $298 \ K$,calculate the emf of the cell.

Consider the following cell reaction:$2Fe_{(s)} + O_{2(g)} + 4H^{+}_{(aq)} \to 2Fe^{2+}_{(aq)} + 2H_2O_{(l)}$; $E^o = 1.67 \ V$At $[Fe^{2+}] = 10^{-3} \ M$,$p(O_2) = 0.1 \ atm$ and $pH = 3$,the cell potential at $25 \ ^oC$ is .............. $V$.

At $298 \ K$,find out the $emf$ for the cell:$Al_{(s)} | Al^{+3} (0.1 \ M) || Fe^{+2} (0.001 \ M) | Fe_{(s)}$Given: $E^o_{Al^{+3}/Al} = -1.66 \ V$ and $E^o_{Fe^{+2}/Fe} = -0.44 \ V$.

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Consider the following cell reaction:
$2Fe_{(s)} + O_{2(g)} + 4H^{+}_{(aq)} \to 2Fe^{2+}_{(aq)} + 2H_2O_{(l)}$; $E^o = 1.67 \ V$
At $[Fe^{2+}] = 10^{-3} \ M$,$p(O_2) = 0.1 \ atm$ and $pH = 3$,the cell potential at $25 \ ^oC$ is .............. $V$.

At $298 \ K$,find out the $emf$ for the cell:
$Al_{(s)} | Al^{+3} (0.1 \ M) || Fe^{+2} (0.001 \ M) | Fe_{(s)}$
Given: $E^o_{Al^{+3}/Al} = -1.66 \ V$ and $E^o_{Fe^{+2}/Fe} = -0.44 \ V$.