Question

Consider a cell composed of two half cells:

1. \[\ce{Cu_{(s)} | Cu{^{2+}_{(aq)}}}\] and
2. \[\ce{Ag_{(s)} | Ag{^+_{(aq)}}}\].

Calculate: (a) the standard cell potential, and (b) the cell potential when [Cu²⁺] is 2M and [Ag⁺] is 0.05 M.

Given: \[\ce{E^{\circ}_{{Cu^{2+}/{Cu}}}}\] = +0.33 V, \[\ce{E^{\circ}_{{Ag^{+}/{Ag}}}}\] = +0.80 V, R = 8.31 J K⁻¹ mol⁻¹, F = 96500 C mol⁻¹.

Answer 1

We are given a galvanic cell composed of:

1. \[\ce{Cu_{(s)} | Cu{^{2+}_{(aq)}}}\] and
2. \[\ce{Ag_{(s)} | Ag{^+_{(aq)}}}\].

And

\[\ce{E^{\circ}_{{Cu^{2+}/{Cu}}}}\] = +0.33 V, 

\[\ce{E^{\circ}_{{Ag^{+}/{Ag}}}}\] = +0.80 V,

R = 8.31 J K⁻¹ mol⁻¹,

F = 96500 C mol⁻¹

We know that

a. \[\ce{E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}}\]

= 0.80 − 0.33

= 0.47 V

b. Cu²⁺ = 2.0 M    ...(given)

Ag⁺ = 0.05 M

Balanced redox reaction:

\[\ce{Cu_{(s)} + 2 Ag{^{+}_{(aq)}} −> Cu{^{2+}_{(aq)}} + 2Ag_{(s)}}\]

`Q = 2.0/(0.05)^2`

= `2.0/0.0025`

= 800

Use the Nernst equation:

E = `E_"cell"^circ - (RT)/(nF) log Q`

= `0.47 - (8.31 xx 298)/(2 xx 96500) log(800)`

= 0.47 − 0.01284 × log(800)

= 0.47 − 0.01284 × 6.6846

= 0.47 − 0.0858

= 0.384 V