1.

How many change in electron and electric charge in balance half reduction reaction for following redox reaction on left hand side ? Cr^(3+)+ClO_(3)^(-)toClO_(2)^(-)+Cr_(2)O_(7)^(2-)

Answer»

`6,-3`
`6,-2`
`6, 3`
`6,-4`

Solution :Reduction HALF reaction :
`UNDERSET((+5))(ClO_(3)^(-))tounderset((+3))(ClO_(2)^(-))`
`ClO_(3)^(-)+2e^(-)underset(2H^(+))toClO_(2)^(-)+H_(2)O`
Number of `e^(-)=2xx3=6`
CHARGES = 3
Balance above reaction multiply by 3 for completion.


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