1870 Rhode Island gubernatorial election
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The 1870 Rhode Island gubernatorial election took place on April 6, 1870, in order to elect the governor of Rhode Island. Republican candidate and incumbent governor Seth Padelford won his second one-year term as governor over Democratic candidate Lyman Pierce.
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April 6, 1870
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 County results Padelford: 50â60% 60â70% | |||||||||||||||||
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Background
In the 1869 Rhode Island gubernatorial election, Lieutenant Governor of Rhode Island Seth Padelford defeated Democratic nominee Lyman Pierce with 68 percent of the vote.[1][2] Pierce was a perennial candidate who had first sought the governorship in 1866 against Republican incumbent Ambrose Burnside.[3][4]
Candidates
- Seth Padelford (Republican), was the incumbent governor. He had previously served in the Rhode Island House of Representatives for two years, ran and lost in the 1860 Rhode Island gubernatorial election as the Republican nominee, and was the Lieutenant Governor of Rhode Island from 1863 to 1865 under James Y. Smith.[2]
- Lyman Pierce, (Democrat), had been nominated in the previous year's election to become the governor of Rhode Island.[5]
Election
Padelford won the election over Lyman by around 25 percent of the vote. He swept all five of the states counties.[3]
| Party | Candidate | Votes | % | |
|---|---|---|---|---|
| Republican | Seth Padelford | 10,337 | 62.15 | |
| Democratic | Lyman Pierce | 6,295 | 37.85 | |
| Total votes | 16,632 | 100.00 | ||
| Republican hold | ||||