What is the effect on the emf of the cell shown in Figure 20.9, which has...
What is the effect on the emf of the cell shown in Figure 20.9, which has the overall reaction Zn(s)+2H+(aq)Zn(s) +2 H+(aq) −→−Zn2+(aq)+H2(g),→ Zn2+(aq) + H2(g), for each of the following changes?
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The pressure of the H2H2 gas is increased in the cathode half-cell.
-
Zinc nitrate is added to the anode half-cell.
-
Sodium hydroxide is added to the cathode half-cell, decreasing [H+].[H+].
-
The surface area of the anode is doubled.
Can someone please explain why b is emf decreases? Why would these ions have an effect on the voltaic system?
Homework Answers
Answer
a) Decrease
b) Decrease
c) Decrease
d) No effect
Explanation
Zn(s) + 2H+ (aq) -------> Zn2+(aq) + H2(g)
Reaction quotient, Q = [Zn2+ ]× PH2/ [H+]
number of electron transfer, n = 2
Nernst equation at 25℃ is
Ecell = E°cell - (0.0592V/n)logQ
From this equation we know that , cell potential depends on Q
If concentration Zn2+ is icreased , cell potential would decresae
Zn2+ concentration is increased by adding zinc nitrate to anode half -cell
So, cell emf decreased by adding zinc nitrate solution to anode half cell.
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