What are the half-reactions for a galvanic cell with Ni and Mg electrodes? A. Ni(s) → Ni2+(aq) + 2e and Mg2+(aq) + 2e → Mg(s)
B. Ni2+(aq) + 2e → Ni(s) and Mg(s) – Mg2+(aq) + 2e
C. Ni(s) Ni2+(aq) + 2e and Mg(s) → Mg2+(aq) + 2e
D. Ni2+(aq) + 2e → Ni(s) and Mg2+ (aq) + 2e → Mg(s)

half-reactions

cathode(reduction) : Ni2+(aq) + 2e → Ni(s)

anode(oxidation) : Mg (s) → Mg²⁺ (aq) + 2e−

Further explanation

Given

Ni and Mg reaction

Required

the half-reactions

Solution

we determine which is the more reduced / oxidized of the two elements by looking at the voltaic series or the standard potential value

In voltaic series  

Li-K-Ba-Ca-Na-Mg-Al-Mn- (H2O) -Zn-Cr-Fe-Cd-Co-Ni-Sn-Pb- (H) -Cu-Hg-Ag-Pt-Au  

The more to the left, the metal is more reactive (easily release electrons) and the stronger reducing agent  

The more to the right, the metal is less reactive (harder to release electrons) and the stronger oxidizing agent

If we look at the Mg metal located to the left of the Ni metal, so the Mg metal is more easily oxidized and can reduce the Ni metal which is on the right

Mg oxidation

Mg(s) – Mg2+(aq) + 2e

Ni reduction

Ni2+(aq) + 2e → Ni(s)

Heat is involved in chemical change in two ways; chemical changes either result in heat being added to or removed from an object, and chemical changes require a certain amount of heat before they can take place.

  remember that hf is only the energy of one photon. not two. not three. one.  1) 5.12 * 10^3 photons  find the energy of one photon, then determine how many photons are needed to reach the sensitivity.  e = hf  = (6.626 * 10^-34 js)(5.95 * 10^19 hz)  = 3.94247 * 10^-14 j per photon  (2.02 * 10^-10 j) / (3.94247 * 10^-14 j/photon) = 5123.69149 photons  round to three sig figs.  2) 5.77 * 10^-17 j  energy of pulse = energy per photon * number of photons  = hfn  = (6.626 * 10^-34 js)(1.66 * 10^13 hz)(5250)  = 5.774559 * 10^-17 j  round to three sig figs.  3) 4.07 * 10^8 kj/mol  10^12 pm = 1 m  c = fλ  f = c/λ  e = hf  = hc/λ  = [(6.626 * 10^-34 js)(3.00 * 10^8 m/s) / (2.94 * 10^-12 m)](1 kj/1000 j)(6.02 * 10^23 photons/mol)  = 40702571.4 kj/mol  round to three sig figs.  4) 9.94 * 10^-1 m  e = hc/λ  λ = hc / e  = (6.626 * 10^-34 js)(3.00 * 10^8 m/s) / (2.00 * 10^-25 j)  = 0.9939 m  round to three sig figs.  5) 9.06 * 10^3 photons  energy of pulse = energy per photon * number of photons  number of photons = energy of pulse / energy per photon  = energy of pulse / (hf)  = (3.87 * 10^-13 j) / [(6.626 * 10^-34 js)(6.45 * 10^16 hz)]  = 9055.23695 photons  round to three sig figs.  6) 2.78 * 10^-15 j  this is problem 1, except in reverse.  sensitivity = minumum number of photons * energy per photon  = (5930)(6.626 * 10^-34 js)(7.077 * 10^14 hz)  = 2.78070758 × 10^-15 j  round to three sig figs.