Question

Which is the correct value for the standard potential for a gold-rhodium voltaic cell?
The standard half cell potential at $298K$ are:
$R{h}^{3+}\left(aq\right)+3e^{-}\to Rh\left(s\right);E_{red}^o=0.76V$
$A{u}^{+}\left(aq\right)+e^{-}\to Au\left(s\right);E_{red}^o=1.69V$

• A. $2.45V$
• B. $0.93V$
• C. $5.83V$
• D. $4.31V$

Answer 1

The correct option is B $0.93V$

The standard reduction potential of the half cells are given.
$E_{R{h}^{3+}\left(aq\right)/Rh\left(s\right)}^{\circ }=0.76V$
$E_{A{u}^{+}\left(aq\right)/Au\left(s\right)}^{\circ }=1.69V$

Metal which having higher reduction potential will undergo reduction and metal having lower reduction potential will undergo oxidation.

So, $Au$ will undergo reduction and act as cathode and $Rh$ will undergo oxidation and act as anode.

The standard cell potential ($\Delta E^o$) of a galvanic cell can be calculated by considering the standard reduction potentials of the two half cells $E^o$.
$E_{cell}^{\circ }=\text{SRP of substance reduced}-\text{SRP of substance oxidised}$
(SRP - Standard reduction potential)

$E_{cell}^{\circ }=E_{cathode\left(red\right)}^{\circ }-E_{anode\left(red\right)}^{\circ }$

$E_{cell}^o=(1.69-0.76)V=0.93V$