Calculate the cell potential, e, for the following reactions at 26.29 °c using the ion concentrations provided. then, determine if the cells are spontaneous or nonspontaneous as written. standard reduction potentials (e°red) may be found here.'1) pt(s)+fe2+(aq)\rightleftharpoonspt2 + (aq)+fe(s)[fe2+]=0.0066m [pt2+]=0.057me= ? v2) cu(s)+2ag+(aq)\rightleftharpoonscu2 +(aq)+2ag(s)[cu2+]=0.013m [ag+]=0.013me= ? v3) co2+(aq)+ti3+(aq)\rightleftharpoons co3+(aq)+ti2+(aq)[co2+]=0.050m [co3+]=0.030m[ti3+]=0.0055m [ti2+]=0.0110me=? vcalculate the cell potential for the following reaction as written at 25.00 °c, given that [mg2 ] = 0.796 m and [sn2 ] = 0.0170 m. standard reduction potentials can be found here. mg(s)+sn2+(aq)\rightleftharpoonsmg2 +(aq)+sn(s)e=? v

The Nernst equation is used to obtain the cell potential under standard conditions from the cell potential under nonstandard conditions.

Pt(s) + Fe^2+(aq) > Pt^2+ (aq) + Fe(s)

We have the following information from the question;

[Fe2+] = 0.0066M,   [Pt2+] = 0.057M

The standard reaction potential is; 1.18 V - (-0.44V) = 1.62 V

Using Nernst equation;

Ecell = E°cell - 0.0592/n log Q

Where;

Q = [Pt2+] /[Fe2+] =  0.057M/ 0.0066M = 8.636

Ecell = 1.62 V - 0.0592/2 log(8.636)

Ecell = 1.59 V

Cu(s) + 2Ag+(aq) > Cu2+(aq) + 2Ag(s)

The standard reaction potential is; 0.80 V - 0.34 V = 0.46 V

Q = [Cu2+]/[Ag+]^2 = 0.013/0.013^2 = 76.92

Ecell =  0.46 V - 0.0592/2 log(76.92)

Ecell = 0.40 V

Co2+(aq) + Ti3+(aq) > Co3+(aq) + Ti2+(aq)

The standard reaction potential of the reaction is; 1.92 V  - (-0.37) = 2.29 V

Q = [Ti2+] [Co3+]/[Co2+] [Ti3+] = [0.0110] [0.030]/[0.050] [0.0055] = 0.00033/0.000275 = 1.2

Ecell = 2.29 V - 0.0592/1 log(1.2)

Ecell = 2.28 V

Mg(s) + Sn2+(aq) > Mg2+(aq) + Sn(s)

The standard reaction potential of the reaction is; (-0.14 V) - (-2.37 V) = 2.23 V

Q = [Mg^2+]/[Sn^2+] = [0.796 M]/[0.0170 M] = 46.8

Ecell = 2.23 V - 0.0592/2 log(46.8)

Ecell = 2.18 V

Each of the cell is spontaneous as written since Ecell in each case is positive.

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I will work only one of the listed equations ... you follow the given example for the remaining reactions. Thank you :-)

Rxn 1: Pt°(s) + Fe⁺²(aq) ⇄ Pt⁺²(aq) + Fe°(s)

a) E(Pt⁺²/Fe°) = - 1.668v

b) Process is Non-spontaneous if E(cell) < 0

Explanation:

Pt°(s) + Fe⁺²(aq) ⇄ Pt⁺²(aq) + Fe°(s) ⇔

Pt°(s)|Pt⁺²[0.057M]║Fe⁺²[0.006M]|Fe°(s)

As written, Pt° is shown undergoing oxidation with Fe⁺² undergoing reduction. Applying the reduction potentials to the analytical equations for E(cell) and ΔG(cell) gives E(Pt/Fe⁺²) < 0 and ΔG(Pt/Fe⁺²) > 0 which indicate a non-spontaneous process. The following supports this conclusion.

E°(Fe⁺²) = -0.44v

E°(Pt⁺²) = +1.20v

E°(Pt/Fe⁺²) =E°(Redn) - E°(Oxidn) =E°(Fe⁺²) - E°(Pt⁺²)

= -0.44v - (+1.20v) = - 1.64v

[Fe⁺²] = 0.0066M

[Pt⁺²] = 0.057M

n = electrons transferred = 2

E(nonstd) = E°(std) - (0.0592/n)logQ);

Q = [Pt⁺²]/[Fe⁺²]

= -1.64v - (0.0592/2)log[0.057M]/[0.006M]v = -1.668v

Also, if ΔG(cell) > 0 => indicates non-spontaneous process

ΔG(Pt/Fe⁺²) = - nFE = -(2)(96,500Coulombs)((-1.664v) > 0 Kj => nonspontaneous rxn. (1 Coulomb-volt = 1 Kilojoule)

b. acid is the correct answer @jayxboof all social media

acid : acids are those which accepts an electron-pairs.

base : bases are those which donates ab electron-pairs.

alkali : it is a base that dissolves in water. it is basically a base and the ph greater than 7.

hydroxide ion : it is a poly-atomic ion which is negatively charged.

1. a)c-14 has two more protons

2. d)they have different numbers of protons, neutrons, and valence electrons; they are all gases at room temperature.

3. c) iron and nickel.