The standard reduction potentials for the following half-cell reactions are, \(2H_2O_{(l)} \rightarrow O_{2(g)} + 4H^+_{(aq)}...
2) \(2H_2O_{2(l)} \rightarrow 2O_{2(g)} + 4H^+_{(aq)} + 4e\), \(E^0 = -0.68V\)
- A) -0.66V
- B) -1.23V
- C) +0.554V
- D) +1.91V
Correct Answer: C) +0.554V
Explanation
The question provides two half-cell reactions with their respective standard reduction potentials:
1) \(2H_2O_{(l)} \rightarrow O_{2(g)} + 4H^+_{(aq)} + 4e\), \(E^0 = -1.23V\)
2) \(2H_2O_{2(l)} \rightarrow 2O_{2(g)} + 4H^+_{(aq)} + 4e\), \(E^0 = -0.68V\)
To determine the standard cell potential (\(E^0_{cell}\) for the overall reaction, we first need to identify the oxidation and reduction half-reactions. Looking at the given reactions, we can see that the first reaction has a more negative standard reduction potential, which means it is more likely to undergo oxidation. So, we reverse the reaction:
\(O_{2(g)} + 4H^+_{(aq)} + 4e \rightarrow 2H_2O_{(l)}\), \(E^0 = +1.23V\)
Now, we have the oxidation half-reaction and the reduction half-reaction:
Oxidation: \(O_{2(g)} + 4H^+_{(aq)} + 4e \rightarrow 2H_2O_{(l)}\), \(E^0 = +1.23V\)
Reduction: \(2H_2O_{2(l)} \rightarrow 2O_{2(g)} + 4H^+_{(aq)} + 4e\), \(E^0 = -0.68V\)
To find the standard cell potential (\(E^0_{cell}\), we add the standard reduction potentials of the oxidation and reduction half-reactions:
\(E^0_{cell} = E^0_{ox} + E^0_{red} = (+1.23V) + (-0.68V) = +0.55V\)
Thus, the standard cell potential for the overall reaction is approximately+0.554V (Option C).