According to the half-reaction table,
`Sn^(2+) + 2e^(-) rightarrow Sn` `E_(@) = -0.14V`
`Mn^(2+) + 2e^(-) rightarrow Mn` `E^(@) = -1.03V`
Which species in the better oxidizing agent?

According to the standard reduction potentials: Pb^(2+)(aq) + 2e^(-) rightarrow Pb(s) E^(@) = -0.13V Fe^(2+)(aq) + 2e^(-) rightarrow Fe(s) E^(@) = -0.44 Zn^(2+)(aq) + 2e^(-) rightarrow Zn(s) E^(@) = -0.76V Which species will reduce Mn^(3+) to Mn^(2+) [E^(@) = 1.51V but will NOT reduce Cr^(3+) to Cr^(2+) [E^(@) = -0.40V]

Use the standard reduction potentials: Sn^(2+)(aq) + 2e^(-) rightarrow Sn(s) E^(@) = -0.141V Ag^(+)(aq) + e^(-) rightarrow Ag(s) E^(@) = 0.800V To calcultate E^(@) for the reaction: Sn(s) + 2Ag^(+)(aq) rightarrow Sn^(2+)(aq) + 2Ag(s)

For a galvanic cell involving the half-reaction at standard conditions, Au^(3+) + 3e(-) rightarrow Au E^(@) = 1.50V Tl^(+) + e(-) rightarrow Tl E^(@) = -0.34V

If {:(Sn^(2+) + 2e^(-) rarr Sn(s), E^(o) = - 0.14 V),(Sn^(4+)+2e^(-) rarr Sn^(2+),E^(o)= + 0.13 V):} then which of these is true?

Standard potentials (E°) for some half-reactions are given below: Sn^(4+) + 2e rightarrow Sn^(2+) , E^(@) = + 0.15V 2Hg^(2+) + 2e rightarrow Hg_(2)^(2+) , E^(@) = 0.92 V pbo_(2) + 4H^(+) + 2e rightarrow pb^(2+)+ 2H_(2)O , E^(@) = + 1.45 V Based on the above , Which one of the following statement is correct?

Which of the half reactions, when coupled, will make a galvanic cell that will produce the largest voltage under standard conditons? P. Cu^(2+)(aq) + 2e^(-) rightarrow Cu(s) E^(@) = +0.34V Q. Pb^(2+)(aq) + 2e^(-) rightarrow Pb(s) E^(@) = -0.13V R. Ag^(+)(aq) + 3e^(-) rightarrow Al(s) E^(@) = -1.66V

Given below are half-cell reaction: Mn^(2+)+2e^(-) rarr Mn,, E^(@) = -1.18 V 2(Mn^(3+)+e^(-) rarr Mn^(2+)),, E^(@) = +1.51 V The E^(@) for 3Mn^(2+) rarr Mn+2Mn^(3+) will be:

Given below are half-cell reaction: Mn^(2+)+2e^(-) rarr Mn,, E^(@) = -1.18 V 2(Mn^(3+)+e^(-) rarr Mn^(2+)),, E^(@) = +1.51 V The E^(@) for 3Mn^(2+) rarr Mn+2Mn^(3+) will be:

Given below are the half -cell reactions Mn^(2+) + 2e^(-) rarr Mn, E^@ =- 1. 18 V 2 (Mn^(3+) = E^(-) rarr Mn^(2+)). E^@ = + 1.5 V The E^@ for Mn^(2+) rarr Mn+ 2 Mn^(3+) will be.