Question

Emf of the cell

Ni | Ni2+ ( 0.1 M ) || Au3+ (1.0 M) | Au will be

$\mathrm { E } _ { \mathrm { Ni } / \mathrm { Ni } ^ { 2 + } } ^ { 0 } = 0.25$ ${E^{0}}_{Au/Au^{3 +}} = - 1.5V$

• A. 1.75 V
• B.
  • 1. 7795 V
• C.
  • 0.7795 V
• D.
  • 1.7795 V

Answer 1

Answer: (B)

• 1. 7795 V

Answer 2

Cell reaction: $3Ni + 2Au^{+ 3}\overset{\quad\quad}{\rightarrow}3Ni^{+ 2} + 2Au$

$E_{cell} = E^{0}cell - \frac{0.0591}{6}\log\frac{\lbrack Ni^{+ 2}\rbrack^{3}}{\lbrack Au^{+ 3}\rbrack^{2}}$

$= ( 0.25 + 1.5 ) - \frac { 0.0591 } { 6 } \log \frac { ( 0.1 ) ^ { 3 } } { ( 1 ) ^ { 2 } } = 1.75 - \frac { 0.0591 } { 2 } \log ( 0.1 )$

= 1.75 + 0.295 = + 1.7795 V