Use the equipment above, as well as wires and a voltmeter, to construct a galvan

Question

Use the equipment above, as well as wires and a voltmeter, to construct a galvanic cell with the largest possible value for E_cell degree. Write the balanced net ionic equation for the reaction that produces the largest possible value for E_cell degree. Calculate the standard cell potential, E_cell degree, in volts for the thermodynamically favored reaction that occurs in this cell. Which metal is used at the anode, and which is used at the cathode? What would be the voltage of this cell if it was constructed without a salt bridge? Explain. The following thermodynamically favored reaction takes place in an acidified galvanic cell. O_2(g) + 2 H_2S(g) rightarrow 2 S(s) + 2 H_2O(l) What is the half reaction that takes place at the anode? What is the half reaction the takes place at the cathode? Calculate the standard cell potential, E_cell degree. What must the partial pressures of the reactants be in order to produce the voltage in part c? Find the total charge that passes though the cell when 0.128 moles of H_2S are consumed. How much charge passes through a wire carrying 2.5 A of current for 16 minutes?

Explanation / Answer

Question 5

The following spontaneous reaction takes place in an acidified voltaic cell.

O2(g) + 2H2S(g) 2S(s) + 2H2O(l)

a.&b What is the half reaction that takes place at the anode?

The cathode? Note that O goes from an oxidation state of 0 to -2.

Therefore, O is reduced at the cathode (Reduction- Cathode ).

S is oxidized at the anode (Anode--- OXidation).

The relevant half reactions from table are…

O2(g) + 4H+ + 4 e- 2H2O E° = 1.23 V…this will take place at the cathode….

S(s) + 2H+ + 2e- H2S E° = 0.14 V (flip this one)…

S(s) + 2H+ + 2e- H2S E° = -0.14 V…this will take place at the anode…

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c. Calculate the standard cell potential, E° for the reaction in this cell

E=E oxidation + E reduction

=1.23 V+- 0.14V=1.09 V ------answer

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.d. What must the partial pressures of the reactants be in order to produce the voltage in B?

Since you are using standard reduction potentials the partial pressures must be at 1 atm.

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e. What volume of oxygen gas is consumed by the reaction if 0.128 mol of H2S is used?

0.128 mol H2S × (1 mole O2/ 2 mole H2S) = 0.0640 mole O2 …then

PV = nRT… V = nRT /P

= [0.0640 mole O2 × (0.0821 L × atm mol × K ) × 298K ]/ 1 atm = 1.57 L O2

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