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2Cr(+III) - 6e = 2Cr(+VI) | 2 | 1
N(+V) + 2e = N(+III) | 6 | 3
b.
2Cr(+VI) + 6e = 2Cr(+III) | 2 | 1
Sn(+II) - 2e = Sn(+IV) | 6 | 3
proszę :)