Question

Calculate the emf and ΔG° for the cell reaction at 25°C: 

\begin{array}{cl}{\mathrm{Zn}(s) | \mathrm{Zn}_{(a q)}^{2+} \|} & {\mathrm{Cd}_{(a q)}^{2+} | \mathrm{Cd}_{(s)}} \\ \ce{(0 \cdot 1 \mathrm{M})} & \ce{(0 \cdot 01 \mathrm{M})}\end{array}

Given E° Zn²⁺/Zn = – 0-763 and E°Cd²⁺/Cd = -0.403

Answer 1

The given electrochemical cell is

\begin{array}{cl}{\mathrm{Zn}(s) | \mathrm{Zn}_{(a q)}^{2+} \|} & {\mathrm{Cd}_{(a q)}^{2+} | \mathrm{Cd}_{(s)}} \\ \ce{\underset{\text{Anode}}{(0 \cdot 1 \mathrm{M})}} & \ce{\underset{\text{Cathode}}{(0 \cdot 01 \mathrm{M})}}\end{array}

`"T" = 25^circC`

`=25+273`

`=298 "K"`

`"E"_"Zn"^circ = - 0.763  "V","E"_"Cd"^circ = - 0.403"V"`

The cell reactions are

\[\ce{Zn_{(s)}->{\mathrm{Zn}_{(aq)}^{2+}} + 2{e^-}\text{(At anode)}}\]

\[\ce{{\mathrm{Cd}_{(aq)}^{2+}} + 2{e^-}->{Cd}_{(s)}\text{(At cathode)}}\]

\[\ce{Zn_{(s)}+{\mathrm{Cd}^{2+}_{(aq)}}->{\mathrm{Zn}_{(aq)}^{2+}} + {\mathrm{Cd}_{(s)}\text{(Cell reation)}}}\]

n = 2

Nernst equation for the cell reaction is 

`="E"_(cell)^circ + (2.303"R""T")/("n""F") log  (["Cd"^(2+)])/(["Zn"^(+2)])`

`=(-0.4.3+0.763)+0.0591/2 log (0.01/0.1)`

`=0.360+0.0591/2 log 10^-1`

`=0.360-0.0591/2 log 10`

`"E"_("cell")= 0.360-0.0591/2xx1`

`= 0.360-0.0296`

`=0.360-0.030`

`"E"_("cell") = 0.330"V"`

`"E"_("cell") = -0.403 + 0.763 = 0.360"V"`

`triangleG^circ = -"n""F""E"_(cell)^circ`

`=-2xx96500xx0.360`

`=-0.72xx96500`

`=-72xx965`

`=-69480"J"`

`triangleG^circ = -63.48"k""J"`