Answer to Question #286579 in General Chemistry for Dos

Question

Given the cell notation, what is [Mn2+] if Ecell reaches 0.055 V? Mn(s) | Mn2+(aq) (0.090 M) || Cd2+(aq) (0.060 M) | Cd(s) at 30 °C

Answer 1

$Cd_{(s)}\to\>Cd^{2+}_{(aq)}+2e^-\>\>\>E^{\theta}=0.40V$

$Mn^{2+}_{(aq)}+2e^-\to\>Mn_{(s)}\>\>\>E^{\theta}=-1.18V$

Total $=1.18-0.40=0.78V$

$E_{cell}=E^0-\frac{(0.0592)}{n}log\>Q$

Let $[Cd^{2+}]$ decreases by $x$

$\implies[Mn^{2+}]$ increases by $x$

$0.055=0.78-\frac{0.0592}{2}\>log\>\frac{0.090+x}{0.060-x}$

Solving

Log $\frac{0.090+x}{0.060-x}=24.493$

$\frac{0.090+x}{0.060-x}=3.112×10^{24}$

Ignoring $x$ and $0.090$ since they are negligible

$1.8672×10^{23}=3.112×10^{24}(x)$

$x=0.06$

$[Mn^{2+}]=0.06+0.09$

$=0.150M$