Question

Calculate the maximum possible electrical work that can be obtained from a galvanic cell under standard conditions at 298K.

\[\ce{Zn|\underset{(aq)}{Zn}^2+ || \underset{(aq)}{Ag}^+ |Ag}\]

Given \[\ce{E°_{(Zn^2+/Zn)} = - 0.76V; E°_{(Ag+/Ag)} = + 0.80 V}\]

Answer 1

Given:

\[\ce{E°_{anode} = - 0.76 V}\]

\[\ce{E°_{cathode} = + 0.80 V}\]

To calculate: W_max

Cell equation: \[\ce{Zn + 2Ag+ -> Zn^{+2} + 2Ag}\]

n = 2

\[\ce{E°_{cell} = E°_{cathode} - E°_{anode}}\]

= 0.80 − (− 0.76)

= + 1.56 V

Now,

\[\ce{W_{electrical} = nFE°_{cell}}\]

= − 2 × 96500 × 1.56

= − 301080 J

Work = 301.08 KJ.

Hence, the maximum possible electrical work in a galvanic cell is 301.08 KJ.