Question

Find Ecell for an electrochemical cell based on the following reaction with [MnO−4]= 1.70 M , [H+]= 1.10 M , and [Ag+]= 0.0150 M . E∘cell for the reaction is +0.880V. MnO−4(aq)+4H+(aq)+3Ag(s)→MnO2(s)+2H2O(l)+3Ag+(aq)

Answer 1

Answer – We are given, [MnO₄^-]= 1.70 M , [H⁺]= 1.10 M , [Ag⁺]= 0.0150 M

MnO₄^- (aq) + 4H⁺ (aq) + 3 Ag(s) -----> MnO₂(s) + 4H₂O(l) + 3Ag⁺(aq), E^ocell = 0.880 V

We know Nernst formula –

Ecell = E^ocell – 0.0592/n * ln K

         = E^ocell – 0.0592/n * ln [Ag⁺]³ / [MnO₄^-][H⁺]⁴

         = 0.880 V – 0.0592 / 3 * ln (0.0150)³ / (1.70)(1.10)⁴

         = 1.15 V

Ecell for an electrochemical cell of the following reaction is 1.15 V