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)?

Pb(s), Fe(s) and Zn(s) will reduce to

Fe(s) and Zn(s) will reduce to

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 value lower than 1.51 V will donate electron to and thus reduces to .

So, Pb(s), Fe(s) and Zn(s) will reduce to .

Metal with value lower than -0.40 V will donate electron to and thus reduces to .

So, Fe(s) and Zn(s) will reduce to .

a boiling water reactor does not boil the water. a pressurized water reactor boils the water faster. a pressurized water reactor is heated directly by the core. a pressurized water reactor does not use fission.