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: The standard reduction potential of X is -1.20 V

Step-by-step 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

Answer 2

`E_(red)^(0)` for X is -1.20 V

Step-by-step explanation:

Oxidation:
`2*`[
`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