A) \[Pt,\frac{1}{2}{{H}_{2}}(1atm)|HCl(1M)\]
B) \[Pt,\frac{1}{2}{{H}_{2}}(1atm)|HCl(2M)\]
C) \[Pt,\frac{1}{2}{{H}_{2}}(1atm)|HCl(0.1M)\]
D) \[Pt,\frac{1}{2}{{H}_{2}}(1atm)|HCl(0.5M)\]
Correct Answer: B
Solution :
: \[{{E}_{cell}}=-\frac{0.591}{1}\log \frac{{{p}_{{{H}_{2}}}}}{[{{H}^{+}}]}\] For [a] \[{{E}_{cell}}==\] For [b] \[{{E}_{cell}}=-0.591\times -0.03010=0.177\] For [c] \[{{E}_{cell}}=-0.591\] For [d] \[{{E}_{cell}}=-0.591\times 0.3010=-0.177\] So only [b] has potential more than zero.
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