Chemistry
11.06.2021 19:05
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Using the standard reduction potentials, pb 2+(aq) + 2e– = pb(s), e° = –0.13 v fe

Using the standard reduction potentials, pb 2+(aq) + 2e– => pb(s), e° = –0.13 v fe 2+(aq) + 2e– => fe(s), e° = –0.44 v zn 2+(aq) + 2e– => zn(s), e° = –0.76 v which metal will reduce mn 3+ to mn 2+ (e° red = +1.51 v) but will not reduce cr 3+ to cr 2+ (e° red = -0.40 v)?
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shymitch32
shymitch32
4,8(64 marks)

Pb(s), Fe(s) and Zn(s) will reduce Mn^{3+} to Mn^{2+}

Fe(s) and Zn(s) will reduce Cr^{3+} to Cr

Explanation:

Standard reduction potential denotes ability to consume electrons from another species.

Hence, higher the standard reduction potential, higher will be the ability to oxidize another species.

Metal with E_{red}^{0} value lower than 1.51 V will donate electron to Mn^{3+} and thus reduces Mn^{3+} to Mn^{2+}.

So, Pb(s), Fe(s) and Zn(s) will reduce Mn^{3+} to Mn^{2+}.

Metal with E_{red}^{0} value lower than -0.40 V will donate electron to Cr^{3+} and thus reduces Cr^{3+} to Cr.

So, Fe(s) and Zn(s) will reduce Cr^{3+} to Cr.

dukeem5270
dukeem5270
4,7(52 marks)

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SUGA

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