Cu(s) + Sn (0.001M) → Cu (0.01M) + Sn(s) The Gibbs free energy change for the above reaction at 298 — Electrochemistry Chemistry Question

Question

Cu(s) + Sn (0.001M) → Cu (0.01M) + Sn(s) The Gibbs free energy change for the above reaction at 298 K is x 10 kJ mol . The value of is __. [nearest integer] [Given: = 0.34 V; = –0.14 V; F = 96500 C mol ] 2+ 2+ –1 –1 –1