Na- - e- - NaThe reaction occurs at-

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Na+ + e- → Na...

Question

$N a^{+} + e^{-} \to N a$
The reaction occurs at:

cathode

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anode

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both A and B

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none of the above

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_{11}^{23} N a +_{1}^{2} H ⟶_{11}^{24} N a +_{1}^{1} H

n (A \cup B \cup C) = n (A) + n (B) + n (c) - n (A \cap B) - n (B \cap C) - n (A \cap C) + n (A \cap B \cap C)

N a - e^{-} \to N a^{+}

N a^{+} (a q) + e^{-} \to N a (s) E^{0} = - 2.71 V

H^{+} (a q) + e^{-} \to \frac{1}{2} H_{2} (g) E^{0} = 0.00 V

(E^{0})

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