Which expression gives the correct value for the standard potential for a gold-rhodium voltaic cell?
Half-reaction $E^{\circ} (V)$
$R h_{(a q)}^{+ 3} + 3 e^{-} \to R h_{(s)}$ $0.76 V$
$A u_{(a q)}^{+} + e^{-} \to A u_{(s)}$ $1.69 V$

Question

Which expression gives the correct value for the standard potential for a gold-rhodium voltaic cell?

Half-reaction	$E^{\circ} (V)$
$R h_{(a q)}^{+ 3} + 3 e^{-} \to R h_{(s)}$	$0.76 V$
$A u_{(a q)}^{+} + e^{-} \to A u_{(s)}$	$1.69 V$

Toppr Admin · Accepted Answer

We know that- metal having higher reduction potential undergoes reduction-so- Au will undergo reduction and Rh will undergo oxidation-E0cell - 1-69 V - 0-76 V

Which expression gives the correct value for the standard potential for a gold-rhodium voltaic cell?Half-reactionE∘(V)Rh+3(aq)+3e−→Rh(s)0.76 VAu+(aq)+e−→Au(s)1.69 V1.69 V−0.76 V1.69 V+0.76 V3(1.69) V+0.76 V3(1.69) V−0.76 V

We know that, metal having higher reduction potential undergoes reduction;so, Au will undergo reduction and Rh will undergo oxidation.E0cell = 1.69 V - 0.76 V

We know that, metal having higher reduction potential undergoes reduction;
so, Au will undergo reduction and Rh will undergo oxidation.
$E_{c e l l}^{0}$ = 1.69 V - 0.76 V