1. For a spontaneous reaction, the Gibbs free energy change must be negative: $\Delta G < 0$

2. The relation between standard free energy change and equilibrium constant is: $\Delta G^o = -RT \ln K$ For spontaneity under standard conditions, $\Delta G^o < 0$, so: $-RT\ln K < 0$ Since $RT>0$, this implies: $\ln K > 0 \Rightarrow K>1$

3. The relation between free energy and cell potential is: $\Delta G^o = -nF E_{\text{cell}}^o$ For a spontaneous cell reaction, $\Delta G^o<0$, hence: $-nF E_{\text{cell}}^o < 0$ Since $nF>0$, we get: $E_{\text{cell}}^o > 0$

4. Therefore, for a spontaneous reaction:

   • $\Delta G$ is negative
   • $K>1$
   • $E_{\text{cell}}^o$ is positive

5. Checking options:

   • A: $-\text{ve}, >1, +\text{ve}$ ✅
   • B: $+\text{ve}, >1, -\text{ve}$ ❌
   • C: $-\text{ve}, <1, -\text{ve}$ ❌
   • D: $-\text{ve}, >1, -\text{ve}$ ❌

Hence, the correct option is A.