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

1.69 V + 0.76 V

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1.69 V - 0.76 V

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

3 (1.69) V + 0.76 V

No worries! We‘ve got your back. Try BYJU‘S free classes today!

3 (1.69) V - 0.76 V

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $1.69 V - 0.76 V$
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

Suggest Corrections

Similar questions

298 K

R h^{3 +} (a q) + 3 e^{-} \to R h (s); E_{r e d}^{o} = 0.76 V

A u^{+} (a q) + e^{-} \to A u (s); E_{r e d}^{o} = 1.69 V

Z n \to Z n^{2 +} + 2 e^{-}; E = + 0.76 V

A u \to A u^{3 +} + 3 e^{-}; E = - 1.42 V

Z n^{2 +}

{Z n}^{+ 2} + {2 e}^{-} ⟶ Z n, E^{o} = - 0.76 V

Z n^{2 +} / Z n, E^{\circ} = - 0.76 V

A g^{+} / A g, E^{\circ} = 0.799 V

{Z n}^{2 +} / Z n

E^{\circ} = - 0.76 V

{A g}^{+} / A g E^{\circ} = 0.799 V

Join BYJU'S Learning Program