Given $E^{\circ} Mn^{7+} / Mn^{2+} = 1.51 \ V$ and $E^{\circ} Mn^{4+} / Mn^{2+} = 1.23 \ V$. Calculate $E^{\circ} Mn^{7+} / Mn^{4+}$. (in $V$)

  • A
    $0.28$
  • B
    $-0.28$
  • C
    $1.70$
  • D
    $0.48$

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For the cell reaction:
$2 Fe^{3+}_{(aq)} + 2 I^{-}_{(aq)} \rightarrow 2 Fe^{2+}_{(aq)} + I_{2(aq)}$
$E^{\ominus}_{cell} = 0.24 \ V$ at $298 \ K$. The standard Gibbs energy $(\Delta_r G^{\ominus})$ of the cell reaction in $kJ \ mol^{-1}$ is:
[Faraday constant $F = 96500 \ C \ mol^{-1}$]

Give the symbolic representation of the following half-cells (electrodes):
$(i)$ $Zn_{(s)} | Zn^{2+}_{(aq)} (1M)$
$(ii)$ $Cu^{2+}_{(aq)} (1M) | Cu_{(s)}$
$(iii)$ $Pt_{(s)} | H_{2(g)} (1 \text{ bar}) | H^+_{(aq)} (1M)$

Identify the representation of the standard hydrogen electrode.

The standard Gibbs energy for the given cell reaction in $kJ \, mol^{-1}$ at $298 \, K$ is $Zn_{(s)} + Cu^{2+}_{(aq)} \to Zn^{2+}_{(aq)} + Cu_{(s)}$,given $E^o = 2 \, V$ at $298 \, K$ [Faraday's constant $F = 96500 \, C \, mol^{-1}$].

Explain: Reduction reaction is possible with higher $E^{\theta}$ value.

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