Question

An electrochemical cell is based on the following two half-reactions:

Ox: Pb(s)→Pb2+(aq, 0.16 M )+2e−

Red: MnO−4(aq, 1.70 M )+4H+(aq, 1.9 M )+3e−→ MnO2(s)+2H2O(l)

Part A Compute the cell potential at 25 ∘C.

Answer 1

Answer:

Cell potential at 25 C is 1.69 V

Explanation:

The given half cell reactions are:

Oxidation (Anode)

$Pb(s)\rightarrow Pb^{2+}(aq, 0.16 M )+2e^{-}-------(1)$

where E°= -0.13 V

Reduction (Cathode):

$MnO_{4}^{-}(aq, 1.70 M )+4H^{+}(aq, 1.9 M )+3e^{-}\rightarrow MnO_{2}(s)+2H_{2}O(l)------(2)$

where E°= +1.51 V

The net reaction is obtained by:

Eq (1) * 3 + Eq(2) * 2

$2MnO_{4}^{-}(aq, 1.70 M )+8H^{+}(aq, 1.9 M )+3Pb(s)\rightarrow  2MnO_{2}(s)+4H_{2}O(l)+3Pb^{2+}(aq, 0.16 M)$

The value of E°(cell) is:

$E_{cell}^{0}=E_{cathode}^{0}+E_{anode}^{0}=1.51-(-0.13)=1.64 V$

The E(cell) at 25 C is calculated based on the nernst equation:

$E_{cell}=E_{cell}^{0}-\frac{0.02568}{n}ln\frac{[Pb^{2+}]^{3}}{[H^{+}]^{8}[MnO_{4}^{-}]^{2}}$

where n = number of electrons = 6

$E_{cell}=1.64-\frac{0.02568}{6}ln\frac{[0.16]^{3}}{[1.9]^{8}[1.70]^{2}}=1.69 V$