Question

Calculate the cell potential at 25oC under the following nonstandard conditions: 2MnO4-(aq) + 3Cu(s) + 8H+(aq) ⟶ 2MnO2(s) + 3Cu2+(aq) + 4H2O(l) [Cu2+] = 0.66M [MnO4-] = 1.69M [H+] = 1.65M Record to the nearest hundredth of a V.

Answer 1

Answer:

1.346 v

Explanation:

1) Fist of all we need to calculate the standard cell potential, one should look up the reduction potentials for the species envolved:

(oxidation) $Cu_{(s)}$ →$Cu^{2+}_{(aq)} +2e$ E°=0.337 v

(reduction) $MnO_{4 (aq)} + 3e + 4H^{+}_{(aq)}$→$MnO_{2 (aq)}+2H_{2}O$ E°=1.679 v

(overall) $2MnO_{4 (aq)}+3Cu_{(s)}$+8H^{+}_{(aq)}→$3Cu^{2+}_{(aq)}+2MnO_{2 (aq)}+4H_{2}O$ E°=1.342 v

2) Nernst Equation

Knowing the standard potential, one calculates the nonstandard potential using the Nernst Equation:

$E=E^{0} -\frac{RT}{nF}Ln\frac{[red]}{[ox]}$

Where 'R' is the molar gas constant, 'T' is the kelvin temperature, 'n' is the number of electrons involved in the reaction and 'F' is the faraday constant.

The problem gives the [red]=0.66M and [ox]=1.69M, just apply to the Nernst Equation to give

$E=1.342 -\frac{298.15*8.314}{6*96500}Ln\frac{[.66]}{[1.69]}=1.346$

E=1.346