Question

6. A galvanic cell is set up as follows: Zn Zn2 (0.01 mol/L) Cu2(0.1 mol/L) Cu. How will the voltage of the cell be affected by each of the following change: (a) increase Cu concentration to Imol/L; (b) decrease Zn2 concentration to 0.000 1 mol/L?

Answer 1

Zu / znŁt Co. 01m0/2) 11 Cu? 6.1m61/2)| Cu Oriclation (anede): Zn(s) — zna tap) + 2e7 Reduction (Cathode): cumi ap) +22 → Cu (s) Coll reaction : Zn(s) + Cut tag? → Zn²+ (aq) + Cu E con = Ecathode - Eanode = Ecuit lou - E24x+/21 = 0.346 -(-0.76V) - H0V 0.0592 V Econ - teen 0.0592 Vlog (2 ) • HON - biogazu ing lo t) HOV - 0.05920 (1) = MOV + 0.0296 V = 1-1296v Thus, be fore changing concentration, Ecev = (1-12960]

2 + 2 -0.0592V (a) [Cu ?*) = I mole Ecen - Eren - o-osa2 log (21) = MOV = 0.0592 cog 20) E LOV-0.0592 (-2) 110V- 0.0592 2 S 1.10 V +0.0592V ] 1.1592 V. Thus, when the cult concertation is increased to twell, the cell intage, Erell is increased to 101592 (5) [2x24) = 0.cool well Ecu = HOV - 0.0592 U log (6.000!). - 1.10 V -0.0592 - 110V + 0.0888V = 1.1888 V Thus, the cell voltage is increased to [1.18887 when the Zn2+ concentration is decreased to 0.0001 molly