Problem: What is the effect on the emf of the cell shown in Figure 20.9 in the textbook, which has the overall reactionZn(s) + 2H+(aq)  →  Zn2+(aq) + H2(g), for each of the following changes?Zinc nitrate is added to the anode compartment.

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We’re asked to determine the effect of adding Zinc nitrate to the anode compartment on the emf of the cell which has the overall reaction: 


Zn(s) + 2H+(aq)    Zn2+(aq) + H2(g)


We will use the Nernst Equation to determine the change in the cell potential with the given conditions. The Nernst Equation relates the concentrations of compounds and cell potential.


Ecell=E°cell-(0.0592n)logQ


Ecell = cell potential under non-standard conditions
 E°
cell = standard cell potential
 n = number of e
- transferred
 Q= reaction quotient = [products]/[reactants] 


The reaction quotient, Q for the overall reaction is:


Q=[products][reactants]

Q=[Zn2+][H+]2

*recall that Q is like an equilibrium expression  solids and liquids are not included
  *the coefficients of the reactants/products will be their exponents in the Q expression


Zinc Nitrate, Zn(NO3)2 dissociates into ions in solution:


Zn(NO3)2 → Zn2+ + 2NO3-


This means that adding Zn(NO3)2  increases [Zn2+] concentration


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Problem Details

What is the effect on the emf of the cell shown in Figure 20.9 in the textbook, which has the overall reaction
Zn(s) + 2H+(aq)  →  Zn2+(aq) + H2(g), for each of the following changes?

Zinc nitrate is added to the anode compartment.

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