Question

consider the reaction between sulfite and a metal anion, X2-, to form the metal, X, and thiosulfate: 2 X2-(aq) + 2 SO32- + 3 H2O(l) → 2 X(s) + S2O32- (aq) + 6 OH- for which Eocell = 0.63. Given that the Eored for sulfite is -0.57 V, calculate Eored for X. Enter your answer to 2 decimal places

Answer 1

Answer:

$E_{red}^{0}$ for X is -1.20 V

Explanation:

Oxidation: $2\times$[$X^{2-}(aq.)-2e^{-}\rightarrow X(s)$]

reduction: $2SO_{3}^{2-}(aq.)+3H_{2}O(l)+4e^{-}\rightarrow S_{2}O_{3}^{2-}(aq.)+6OH^{-}(aq.)$

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overall:$2X^{2-}(aq.)+2SO_{3}^{2-}(aq.)+3H_{2}O(l)\rightarrow 2X(s)+S_{2}O_{3}^{2-}(aq.)+6OH^{-}(aq.)$

So, $E_{cell}^{0}=E_{red}^{0}(SO_{3}^{2-}\mid S_{2}O_{3}^{2-})-E_{red}^{0}(X\mid X^{2-})$

or, $0.63=-0.57-E_{red}^{0}(X\mid X^{2-})$

or, $E_{red}^{0}(X\mid X^{2-})= -1.20$

So, $E_{red}^{0}$ for X is -1.20 V

Answer 2

Answer: The standard reduction potential of X is -1.20 V

Explanation:

For the given chemical equation:

$2X^{2-}(aq.)+2SO_3^{2-}+3H_2O(l)\rightarrow 2X(s)+S_2O_3^{2-}(aq.)+6OH^-$

The half reaction follows:

Oxidation half reaction:  $X^{2-}(aq.)\rightarrow X(s)+2e^-$     ( × 2 )

Reduction half reaction:  $2SO_{3}^{2-}(aq.)+3H_{2}O(l)+4e^{-}\rightarrow S_{2}O_{3}^{2-}(aq.)+6OH^{-}(aq.)E^o=-0.57V$

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

We are given:

$E^o_{cell}=0.63V$

Putting values in above equation, we get:

$0.63=-0.57-E^o_{anode}\\\\E^o_{anode}=-0.57-0.63=-1.20V$

Hence, the standard reduction potential of X is -1.20 V