Given: Cr^3+(aq) + 3e^- Cr(s); E degree = -0.74 V Pb^2+(aq) + 2e^- Pb(s); E degr

Question

Given: Cr^3+(aq) + 3e^- Cr(s); E degree = -0.74 V Pb^2+(aq) + 2e^- Pb(s); E degree = -0.13 V What is the standard cell potential for the following reaction? 2Cr(s) + 3Pb^2(aq) rightarrow 3Pb(s) + 2Cr^3t(aq) A) -0.87V B) 1.83 V C) -0.61V D) 0.61V E) 0.87V Given: Zn^2+(aq) + 2e^- Zn(s); E degree = -0.76 V Cu^2+(aq) + 2e^- Cu(s); E degree = 0.34 V What is the cell potential of the following electrochemical cell at 25 degree C? Zn(s)|Zn^2+(1.0 M) || Cu^2+(0.0010 M) | Cu(s) A) between 0.34 and 0.76 V B) less than 0.42 V C) between 0.00 and 0.76 V D) between 0.76 and 1.10 V E) greater than 1.10 V

Explanation / Answer

18)The cell is Cr/Cr+3 || Pb+2/Pb

Thus E0 cell = SRP of Pb - SRP of Cr

= -0.13 - (-0.74)V

= +0.61V

OPtion D

19) The cell is Zn/Zn+2 || Cu+2/Cu

E0 cell -= 0.34 -(-0.76V)

= 1.10V

For the cell

Zn/Zn+2(1.0M) || Cu+2(0.001M)/Cu

using Nernst equation

Ecell = Eocell - 0.059/2 log [Cu+2]/[zn+2]

= 1.10V - 0.059/2 log 1/10-3

= 1.01V

Thus Option D is right.

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