Calculate the cell potential for $Ni_{(s)} | Ni^{2+}(0.036 \ M) || Co^{2+}(0.018 \ M) | Co_{(s)}$ at $298 \ K$.

  • A
    -$0.029$ \ $V$
  • B
    -$0.059$ \ $V$
  • C
    $0.029$ \ $V$
  • D
    $0.059$ \ $V$

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$Cu_{(s)} | Cu^{+2}(aq, 10^{-3} M) || Ag^{+}(aq, 10^{-5} M) | Ag_{(s)}$
If $E^{o}_{Cu^{+2}/Cu} = +0.34 \ V$
$E^{o}_{Ag^{+}/Ag} = +0.80 \ V$
$E_{cell}$ will be

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The cell potential for $Zn | Zn^{2+}_{(aq)} || Sn^{x+}| Sn$ is $0.801 \ V$ at $298 \ K$. The reaction quotient for the above reaction is $10^{-2}$. The number of electrons involved in the given electrochemical cell reaction is .... (Given $E^{0}_{Zn^{2+}|Zn} = -0.763 \ V, E^{0}_{Sn^{x+}|Sn} = +0.008 \ V$ and $\frac{2.303 \ RT}{F} = 0.06 \ V$)

Calculate the equilibrium constant for the cell reaction at $298 \ K$: $Cu_{(s)} + 2Ag^{+}_{(aq)} \rightarrow Cu^{2+}_{(aq)} + 2Ag_{(s)}$,given $E^{0}_{cell} = 0.46 \ V$.

For the cell reaction $Sn_{(s)} + 2Ag^+_{(aq)} \rightarrow Sn^{2+}_{(aq)} + 2Ag_{(s)}$,what happens when the cell voltage increases?

What is the sign of $\Delta G^o$ and the value of $K$ for an electrochemical cell for which $E_{cell}^o = 0.80 \ V$?

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