Given,for Sn^{4+} / Sn^{2+},standard reductio | vedclass

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Question

(A) The given reaction is $3 Sn^{2+} + 2 Au^{3+} \longrightarrow 3 Sn^{4+} + 2 Au$. In this reaction,$Sn^{2+}$ is oxidized to $Sn^{4+}$ (anode) and $A

Vedclass · Accepted Answer

$+1.35 \ V$

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(A) The given reaction is $3 Sn^{2+} + 2 Au^{3+} \longrightarrow 3 Sn^{4+} + 2 Au$. In this reaction,$Sn^{2+}$ is oxidized to $Sn^{4+}$ (anode) and $Au^{3+}$ is reduced to $Au$ (cathode). The standard reduction potential for the cathode is $E_{	ext{cathode}}^{\circ} = E_{Au^{3+}/Au}^{\circ} = 1.5 \ V$. The standard reduction potential for the anode is $E_{	ext{anode}}^{\circ} = E_{Sn^{4+}/Sn^{2+}}^{\circ} = 0.15 \ V$. The standard cell potential is calculated as: $E_{	ext{cell}}^{\circ} = E_{	ext{cathode}}^{\circ} - E_{	ext{anode}}^{\circ} = 1.5 \ V - 0.15 \ V = 1.35 \ V$.

Given,for $Sn^{4+} / Sn^{2+}$,standard reduction potential is $0.15 \ V$ and for $Au^{3+} / Au$,standard reduction potential is $1.5 \ V$. For the reaction,$3 Sn^{2+} + 2 Au^{3+} \longrightarrow 3 Sn^{4+} + 2 Au$,the value of $E_{\text{cell}}^{\circ}$ is:

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