Question

A galvanic cell at a temperature of 25.0 °C is powered by the following redox reaction: Cu? + (aq) + Zn (s) — Cu(s) +Zn² + (aq) Suppose the cell is prepared with 6.36 M Cu? in one half-cell and 3.80 M Zn²+ in the other. Calculate the cell voltage under these conditions. Round your answer to 3 significant digits. 0u ou X 5 ?

Answer 1

The complete calculation of cell voltage is given as below:-

Reaction is given by :- Cu²+ (aq) + Zn (5) Cu(s) + Zn2+ (92) - 0 oxidation reaction = Zn (s) Zn²(aq) tre, E = +0.76V Reduction reaction > Cu?taq) +28 - Cus), E = +0.34 40 E = Eloxidation) + E(Reduction) where, Et = Total Standard electrode potential. E = +0.76V 70-34V =+lV or [E = 1.10

Also it is given that, [cu2t] = 6.36m and [Zn2+] = 380m T= 25°C = 298K =Temperature Number of electrons transfered = 2 elections. From Nernst's equation, we have E - Eº – 0-0591 log [product) - [Reactant] n

For concentration cell, we have E;= E° -0, n=2 Then from equation ③, we get E=0-01059 1 log [Znet 2 [cu2t) log (3.80m ) or E= -0.02955 on E = -0.029 55 log (0.598) = +0.02955 (022) or E= 6.501 810-3 v So, the cell voltage will be E = 6.501 8103 v