| Election Name: | 1869 Rhode Island gubernatorial election |
| Before Party: | Republican |
| Before Election: | Ambrose Burnside |
| After Party: | Republican |
| After Election: | Seth Padelford |
| Governor | |
| Party Name: | no |
| Popular Vote1: | 7,359 |
| Type: | presidential |
| Party1: | Republican |
| Nominee1: | Seth Padelford |
| Image1 Upright: | .5 |
| Image1: | SethPadelford.jpg |
| Percentage1: | 68.46% |
| Party2: | Democratic |
| Nominee2: | Lyman Pierce |
| Popular Vote2: | 3,390 |
| Percentage2: | 31.54% |
| Map Size: | 225px |
| Election Date: | April 7, 1869 |
| Ongoing: | no |
| Next Year: | 1870 |
| Next Election: | 1870 Rhode Island gubernatorial election |
| Previous Year: | 1868 |
| Previous Election: | 1868 Rhode Island gubernatorial election |
The 1869 Rhode Island gubernatorial election took place on April 7, 1869 in order to elect the governor of Rhode Island.[1] Republican candidate and incumbent governor Seth Padelford won his first one-year term as governor[2] against Democratic candidate Lyman Pierce.[3]