Designing a galvanic cell from two half-reactions A chemist designs a galvanic c

Question

Designing a galvanic cell from two half-reactions A chemist designs a galvanic cell that uses these two half-reactions: half-reaction standard reduction potential Ered =-0.763 V red = + 1.00 V Zn- (aq)+2e Zn(s) 0 Answer the following questions about this cell Write a balanced equation for the half-reaction that happens at the cathode x10 Write a balanced equation for the half-reaction that happens at the anode Write a balanced equation for the overall reaction that powers the cell. Be sure the reaction is spontaneous as written Do you have enough information to calculate the cell voltage under standard conditions? Yes If you said it was possible to calculate the cell voltage, do so and enter your answer V here. Round your answer to significant digits

Explanation / Answer

a)

cathode -> gain of electrons, the higher potential goes here

2HNO2(aq) + 2H+(aq) + 2e- = 2NO(g) + 2H2O(l)

b)

anode --> oxidaiton, loss of e-, invet rxn of Zn

Zn(s) --> Zn2+(aq) + 2e-

c)

Add both equations

2HNO2(aq) + 2H+(aq) + 2e- = 2NO(g) + 2H2O(l)

Zn(s) --> Zn2+(aq) + 2e-

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Zn(s) + 2HNO2(aq) + 2H+(aq) + 2e- = 2NO(g) + 2H2O(l) + Zn2+(aq) + 2e-

cancel e-

Zn(s) + 2HNO2(aq) + 2H+(aq)= 2NO(g) + 2H2O(l) + Zn2+(aq)

c)

Yes, at STD conditions, we do

d)

E°cell = Ecahtode - Eanode = (1) - (-0.763) = 1.763 V

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