The $EMF$ of the cell $Cr|Cr^{3+}(0.1 \ M) \| Fe^{2+}(0.01 \ M)|Fe$ is: $(E^{\circ}_{Cr^{3+}|Cr} = -0.75 \ V, E^{\circ}_{Fe^{2+}|Fe} = -0.45 \ V)$ (in $V$)

  • A
    $0.30$
  • B
    $0.36$
  • C
    $0.26$
  • D
    $0.22$

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Similar Questions

For the cell $Cu_{(s)}|Cu^{2+}_{(aq)}(0.1 \ M) || Ag^{+}_{(aq)}(0.01 \ M)| Ag_{(s)}$,the cell potential $E_{1} = 0.3095 \ V$. For the cell $Cu_{(s)}|Cu^{2+}_{(aq)}(0.01 \ M) || Ag^{+}_{(aq)}(0.001 \ M)| Ag_{(s)}$,the cell potential $= ..... \times 10^{-2} \ V$. (Round off to the Nearest Integer). [Use: $\frac{2.303 \ RT}{F} = 0.059$]

What is the potential of a hydrogen electrode in contact with a solution whose $pH$ is $1$ (in $V$)?

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

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Determine the electrode potential of $Sn^{2+} (0.01 \text{ M}) | Sn(s)$ at $25^\circ \text{C}$ if $E^\circ_{Sn^{2+}/Sn} = -0.136 \text{ V}$. (in $\text{ V}$)

At $298 \ K$ the equilibrium constant for the reaction $M_{(s)} + 2 Ag^{+}_{(aq)} \rightarrow M^{2+}_{(aq)} + 2 Ag_{(s)}$ is $10^{15}$. What is the $E_{cell}^{\ominus}$ (in $V$) for this reaction? $\left(\frac{2.303 RT}{F}\right) = 0.06 \ V$

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