Consider the table of standard reduction potentials shown below.
Half-reaction $E^{\circ}$
$C l_{2} + 2 e^{-} \to 2 C l^{-}$ $1.36 V$
$O_{2} + 4 H^{+} + 4 e^{-} \to 2 H_{2} O$ $1.23 V$
$2 H_{2} O + 2 e^{-} \to H_{2} + 2 O H^{-}$ $- 0.83 V$
$R b^{+} + e^{-} \to R b$ $- 2.93 V$
Use the information from the table and your knowledge of electrochemistry to predict the CORRECT net ionic equation for the reaction that will occur when an aqueous solution of rubidium chloride undergoes electrolysis.

2 R b^{+} + 2 C l^{-} \to 2 R b + C l_{2}

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2 H_{2} O \to 2 H_{2} + O_{2}

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H_{2} + 2 O H^{-} + 2 R b^{+} \to 2 R b + 2 H_{2} O

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4 C l^{-} + O_{2} + 4 H^{+} \to 2 H_{2} O + 2 C l_{2}

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Solution

The correct option is A $2 R b^{+} + 2 C l^{-} \to 2 R b + C l_{2}$
From the given table, we have;
Rb as the lowest reduction potential therefore it undergo oxidation
and $C l_{2}$ have the highest oxidation potential therefore it undergo reduction.
so, the net reaction becomes

$2 R b^{+} + 2 C l^{-} \to 2 R b + C l_{2}$

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Similar questions

2 N a + 2 H_{2} O \to 2 N a O H + H_{2}

2 N a O H + C l_{2} \to N a C l + N a O C l + H_{2} O

4 O H^{-} \to O_{2} + 2 H_{2} O + 4 e^{-}

2 C l^{-} \to C l_{2} + 2 e^{-}

$2 C l^{-} \to C l_{2} + 2 e^{-}$
$2 H_{2} O + 2 e^{-} \to H_{2} + 2 O H^{-}$

Which of the following is correct regarding liberation of electrolytic products?

{2 H}_{2} O + {2 e}^{-} \to H_{2} + 2 {O H}^{-}

{x O H}^{-} \to O_{2} + {2 H}_{2} O + {y e}^{-}

{2 H}_{2} O \to 2 H_{2} + O_{2}

N a C l

2 O H^{-} + C l_{2} \to 2 O C l^{-} + H_{2} O

2 N a + 2 H_{2} O \to 2 N a O H + H_{2}

4 O H^{-} \to O_{2} + 2 H_{2} + 4 e^{-}

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