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Consider the following Galvanic cell:- ...

Consider the following Galvanic cell:-

By what value the cell voltage when concentration of ions in anodic and cathodic compartments both increased by factor of 10 at 298K:

A

`+0.0591`

B

`-0.0591`

C

`-0.1182`

D

0

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The correct Answer is:
C
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The Nernst equation: Consider the following galvanic cell Pb(s) | Pb^(2+)(aq.) || Ag^(2+)(aq.) | Ag(s) (a) What is the quantitative change in the cell voltage on increas-ing the ion concentrations in the anode compartment by a factor of 10 ? (b) What is the quantitative change in the cell voltage on increas-ing the ion concentration in the cathode compartment by a factor of 10 Strategy: The conventional notation of the cell tells us that lead is the anode while silve is the cathode. Therfore, the cell reaction is Pb(s) + 2Ag^(+)(aq.) rarr Pb^(2+)(aq.) + 2Ag(s) The cell potantial (at 25^(@)C ) is given by the Nernst equation, where n = 2 and Q = C_(Pb^(2+))//C_(Ag^(+)) : E_("cell") = E_("cell")^(@)-(0.0592 V)/(n) "logQ" E_("cell")^(@)-(0.0592 V)/(n) "logQ" C_(Pb^(2+))/C_(Ag^(+))^(2)

Write the Nernst equation for the cell reaction in the Daniel cell. How will the E_("cell") be affected when concentration of Zn^(+) ions is increased?

Consider the following relations for emf of a electrochemical cell (i) emf of cell = (Oxidation potential of anode)-(Reduction potential of cathode) (ii) emf of cell = (Oxidation potential of anode)+(Reduction potential of cathode) (iii) emf of cell = (Reduction potential of anode)+(Reduction potential of cathode) (iv) emf of cell = (Oxidation potential of anode)-(Oxidation potential of cathode) Which of the above realtions are correct?

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