Select the single best answer. Consider the following half-reactions: MnO4−(aq) + 8H+(aq) + 5e−→ Mn2+(aq) + 4H2O(l) NO3−(aq) + 4H+(aq) + 3e− → NO(g) + 2H2O(l) Will NO3− oxidize Mn2+ to MnO4− under standard-state conditions? NO3− ions will oxidize Mn2+ to MnO4− under standard state conditions. NO3− ions will not oxidize Mn2+ to MnO4− under standard state conditions.

NO3− ions will not oxidize Mn2+ to MnO4− under standard state conditions.

Explanation:

The reduction potential of NO3^- is +0.96V while the reduction potential of MnO4^- is +1.51V. Hence looking at the reduction potential values stated above, NO3^- will not oxidize Mn^2+ to MnO4^-.

The MnO4^- having a greater reduction potential will be reduced to Mn^2+ while NO3^- will be oxidized to NO.

it would be b. 6co2 + 6h2o --> c6h12o6 + 6o2

answer : third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol

explanation :

let us write down the given equations

equation 1 :

equation 2 :

equation 3 :

let us use hess's law to find out which 2 equations can be added to get the third equation.

let us reverse equation 2 and multiply it by 2. the new equation that we get is,

equation 4 :

let us add equation 1 & 4

the above equation is similar to equation 3 but has different value for dhrxn.

therefore third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol