electrochemistry i submitted this previously but the image got distorted on the

Question

electrochemistry

i submitted this previously but the image got distorted on the upload. hopefully this is better

S. This problem deals with a battery for the overall reaction Zn(s) 2 Ag (aq) The cell is constructed as follows: The silver metal electrode weighs 10.0 g The zinc metal electrode weighs 10.0 g. water, The volume The left compartment contains 10.0 g of silver(I) sullate dissolved in of this solution is 100.0 mL volume of The right compartment contains 10.0 g of zinc sulfate ved in water. The his solution is 100.0 mL A current of96.5 Amps has passed through the battery for 10 sec. (a) What is the concentration in molM of silver ion in the left compartment after this charge has passed? after this (b) What is the concentration in mollL of zinc ion in the right compartment charge has passed? (e) What is the mass of the zine electrode after this charge has passed? The battery continues to run until it is completely dead. (d) How many moles of electrons (total) have passed? (e) What is the concentration in Lof silver ion in the left compartment after this charge has passed?

Explanation / Answer

Write down the half equations:

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

Ag+ (aq) + e- -----> Ag (s)

As per the given reaction, Ag+ is reduced while Zn is oxidized.

Molar mass of Ag(I) sulfate = 311.799 g/mol.

Therefore, moles of Ag(I) sulfate present = (10.0 g)*(1 mol/311.799 g) = 0.0321 mol.

Molar mass of Zn(II) sulfate = 161.47 g/mol.

Moles of Zn(II) sulfate present = (10.0 g)*(1 mol/161.47 g) = 0.0619 mol.

a) Current passed = 96.5 Amps = 96.5 C/s; time = 10 secs.

Therefore, charge passed = (96.5 C/s)*(10 s) = 965 C.

1 Faraday of electricity = 96500 C.

Therefore, 965 C = (965 C)*(1 F/96500 C) = 0.01 F

1 Faraday = 1 mole of electrons.

Therefore, 0.01 F = 0.01 mole of electrons.

As per the balanced equation,

1 mole of electrons = 1 mole of Ag+ reduced.

Therefore, 0.01 mole of electrons = 0.01 mole Ag+ reduced = 0.01 mole Ag+ consumed.

Moles Ag+ left = (0.0321 – 0.01) mole = 0.0221 mol.

Volume of the compartment = 100.0 mL = 0.1 L.

Therefore, concentration of Ag (I) left = (0.0221 mol)/(0.1 L) = 0.221 mol/L (ans).

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