1860 Rhode Island gubernatorial election

1860 Rhode Island gubernatorial election

April 4, 1860
 
Nominee William Sprague IV Seth Padelford
Party Democratic Republican
Alliance Conservative
Popular vote 12,278 10,740
Percentage 53.3% 46.7%

County results
Sprague:      50–60%
Padelford:      50–60%

Governor before election

Thomas G. Turner
Republican

Elected Governor

William Sprague IV
Democratic

A gubernatorial election was held in Rhode Island on April 4, 1860. The Democratic and Conservative candidate William Sprague IV defeated the Republican former member of the Rhode Island House of Representatives from Providence Seth Padelford.

At the Republican state convention, Padelford, a Radical Republican, defeated the incumbent governor Thomas G. Turner. Padelford's nomination precipitated a movement by Conservative Republicans, former Whigs, and Democrats to jointly nominate Sprague as the conservative fusion candidate. Sprague's surrogates portrayed the election as a choice between "Union and Disunion" and charged Padelford with representing "agitation, anarchy," and "black Republicanism." The conservatives used several names during the campaign, including "Young Men's," "Conservative Republican," "Democratic," "Military," "Conservative Union," and "American." Sprague's coalition subsequently became the foundation for the Rhode Island Constitutional Union Party.