Question

A voltaic cell consists of a Pb/Pb2+ half-cell and a Cu/Cu2+ half-cell at 25 ∘C. The initial concentrations of Pb2+ and Cu2+ are 5.30×10−2 M and 1.70 M, respectively.

Part A- What is the initial cell potential? Express your answer using two significant figures.

Part B- What is the cell potential when the concentration of Cu2+ has fallen to 0.230 M ? Express your answer using two significant figures.

Part C- What are the concentrations of Pb2+ and Cu2+ when the cell potential falls to 0.36 V? Enter your answers numerically separated by a comma. Express your answers using two significant figures.

Answer 1

Solution : - Given datas- pbat a 5-30x10 ²M co²t = 1.70 M. PART: A pool phe case, (São *02) Il company (1 70)/eve) Epost P -0112.67 E Coat / cu =+0 • 34 V = 25°c +273 = 298 K Neomast Equition : - f cell = 0.34 - (0:126) 50.466V Fede z Ecele - moi co .: celle Paris i pbcss teuat pbt + co) - 6-466 2-303 RT 0.0530 2F 67 0.0530 2 0466 0.0541 log 0.059/ ... -0.466 - oss. 2- 1.

Fcell = 0.5low Ecell = 0.3lov / Ran: Eqution - - 1 mol cult produce 1 mol pb²+ Let, se a change in come of both 30/.0.37 .= 0.466 – og lang (00053 + x) (107-22 slog (0.053 + 2 O.166 – 0:37 7107 -2] = 0.0591 0.0591 0.053 tx) 7037 0.466 - = 2249 0.053 ta =774019 R- te 0.053 +2 = 3016. 12 -1774.192 1975.19x = 3016.12 -0.053 3015. 947 .3 1.64 11750193 ze = 107 i ph2+] = 0.053 +10699 rcost] = 107 -1.699 1.752 M = ooool in

& concentration of cost 0.230m of chang 17 -0.230 = 1047 et concentration of pb I 0.053 t1.47 1.523 L Go - Ecelleme 0.0591 - 1.523 - 503583 V 01230 mm 2 Ecell 0.3583 V